Standard Forms of Equation of Circles | Diameter Form, Polar Form, Parametric Form of Circle | Equation of Circle Touching X-axis, Y-axis, Both axis

Define Circle

A circle is the locus of points which moves in a plane such that its distance from a fixed point is always constant. The fixed point is called the ‘centre’ while the fixed distance is called the ‘radius’.

Equation of a Circle When the Centre is Origin

Consider an arbitrary point P(x, y) on the circle. Let ‘r’ be the radius of the circle which is equal to OP.

We know that the distance between the point (x, y) and origin (0,0)can be found using the distance formula which is equal to-

OP = r

√[(x-0)²+ (y-0)²]= r

x²+ y²= r²

Therefore, the equation of a circle, with the center as the origin is,

x²+y² = r²

Where “r” is the radius of the circle.

Example : Consider a circle whose center is at the origin and radius is equal to 8 units.

Solution:

Given: Centre is (0, 0), radius is 8 units.

We know that the equation of a circle when the center is origin:

x²+ y² = r²

For the given condition, the equation of a circle is given as

x² + y² = 8²

x² + y²= 64, which is the equation of a circle

Equation of a Circle When the Centre is not an Origin

Let C(h, k) be the centre of the circle and P(x, y) be any point on the circle.

Therefore, the radius of a circle is CP.

By using distance formula,

(x-h)² + (y-k)² = CP²

Let radius be ‘r’.

Therefore, the equation of the circle with center (h, k)and the radius ‘a’ is,

(x-h)²+(y-k)² = r²

which is called the standard form for the equation of a circle.

In general, if the centre of the circle is represented with (h, k) and the radius of the circle is ‘r’ then the equation of the circle is,

(x-h)² + (y-k)² = r²

This equation represents all the points that are at r distance from the point (h, k) in the 2-D coordinate plane.

This is the standard form of the equation. Thus, if we know the coordinates of the center of the circle and its radius as well, we can easily find its equation.

Example.1 : If the center and the radius of the circle are (3, 5) and 7 units respectively then find the equation of the circle.

Solution: 

Centre of the circle is (3, 5) comparing with (h, k) we get h = 3 and k = 5

Similarly, radius of the circle is 7 = r

Equation of the circle is (x – h)² + (y – k)² = r² 

(x – 3)² + (y – 5)² = 7² 

This is the required equation of the circle.

Example.2 : What is the equation of the circle if the point (1,2) is the center of the circle and radius is equal to 4 cm ?

Solution :

Centre of the circle is (1, 2) comparing with (h, k) we get h = 1 and k = 2

Similarly, radius of the circle is 4 = r

Equation of the circle is (x – h)² + (y – k)² = r² 

(x-1)²+(y-2)² = 4²

(x²−2x+1)+(y²−4y+4) =16

X²+y²−2x−4y-11 = 0

Example 3: Find the equation of the circle whose center is (3,5) and the radius is 4 units.

Solution:

Here, the center of the circle is not an origin.

Therefore, the general equation of the circle is,

(x-3)² + (y-5)²  = 4²

x² – 6x + 9 + y² -10y +25 = 16

x² +y² -6x -10y + 18 =0

General Equation of a Circle

The general form of the equation of a circle is represented as,

x² + y² + 2gx + 2fy + c = 0

where,

• g, f, and c are constant values

Using this general form of the equation we can easily find the centre and radius of the circle.

Converting the general form of the equation x² + y² + 2gx + 2fy + c = 0, into the standard form

x² + y² + 2gx + 2fy + c = 0

Adding f², g², and -c on both sides,

x² + y² + 2gx + 2fy + c + f² + g² -c = f² + g² -c

x² + 2gx + g² + y² + 2fy + f² = f² + g² -c

(x + g)² + (y + f)² = [√(f² + g² -c)]²

comparing with (x-h)² + (y-k)² = r²

we get,

• centre = (h, k) = (-g, -f)
• radius = r = √(f² + g² -c)

Features of General Equation of Circle

x² + y² + 2gx + 2fy + c = 0, represents the general equation of a circle with centre (−g, −f) and radius (r): r² = g² + f²− c.

Some of the important results which we deduce from the general equation of the circle are,

• If g² + f² > c, then the radius of the circle is real.
• If g² + f² = c, then the radius of the circle is zero which tells us that the circle is a point which coincides with the centre. Such type of circle is called a point circle
• g² + f² <c, then the radius of the circle becomes imaginary. Therefore, it is a circle having a real centre and imaginary radius. 

What is Equation of Point Circle?

A point circle is the circle whose radius is zero. The equation of the point circle is,

(x-a)² + (y-b)² = 0

Important Points

The general form of the equation of the circle is x² + y² + 2gx + 2fy + c = 0. Important Points of the equation of the circle are,

• It is quadratic in both x and y.
• Coefficient of x² = y². (It is advisable to keep the coefficient of x² and y² unity)
• There is no term containing xy i.e., the coefficient of xy is zero.
• It contains three arbitrary constants viz. g, f and c.

Example 1: Equation of a circle is x²+y²−12x−16y+19=0. Find the center and radius of the circle.

Solution:

Given equation is of the form x²+ y² + 2gx + 2fy + c = 0,

2g = −12, 2f = −16,c = 19

g = −6,f = −8

Centre of the circle is (6,8)

Radius of the circle = √[(−6)² + (−8)² − 19 ]= √[100 − 19] =

=  √81 = 9 units.

Therefore, the radius of the circle is  9 units.

Position of a Point w.r.t. to Circle

Let the circle be x² + y² + 2gx + 2fy + c = 0 and P(x₁, y₁) be the point.

R – radius

CP > R , {point lie outside}

CP = R , {on the curve}

CP < R , {inside the curve}

Equation of Circle Touching x-axis

Suppose the circumference of the circle touches the x-axis at point (a, 0) and (a, r) is the centre of the circle with radius r. 

We know that if a circle touches the x-axis, then its y-coordinate of the centre of the circle is equal to the radius r, this condition is shown in the figure below,

Take any general point (x, y) on the circumference of the circle then its equation is given as,

Distance between (x, y) and (a, r) = Radius(r)

√[(x -a)² + (y-r)²] = r

Squaring both sides we get,

(x -a)² + (y-r)² = r²

This is the equation of the circle touching the x-axis.

Equation of Circle Touching Y-axis

Suppose the circumference of the circle touches the y-axis at point (0, b) and (r, b) is the centre of the circle with radius r. 

We know that if a circle touches the y-axis, then its x-coordinate of the centre of the circle is equal to the radius r, this condition is shown in the figure below,

Take any general point (x, y) on the circumference of the circle then its equation is given as,

Distance between (x, y) and (r, b) = Radius(r)

√[(x -r)² + (y-b)²] = r

Squaring both sides we get,

(x -r)² + (y-b)² = r²

This is the equation of the circle touching the y-axis.

Equation of Circle Touching Both X and Y Axes

Suppose the circumference of the circle touches the x-axis at point (r, 0)and the y-axis at point (0, r) and (r, r) is the centre of the circle with radius r. 

We know that if a circle touches both x and y-axis, then its x-coordinate and y-coordinate of the centre of the circle are equal to the radius r, this condition is shown in the figure below,

Take any general point (x, y) on the circumference of the circle then its equation is given as,

Distance between (x, y) and (r, r) = Radius(r)

√[(x -r)² + (y-r)²] = r

Squaring both side we get,

(x -r)² + (y-r)² = r²

This is the equation of the circle touching both the axes i.e. x-axis and y-axis.

Equation of Circle Passes through the origin with centre (h, k)

General Equation of circle is :

(x-h)²+(y-k)² = r²

Circle passes through origin O(0, 0)

(0-h)²+(0-k)² = r²

r² = h² + k² 

Put value of r² in general equation, we get

(x-h)²+(y-k)² = h² + k² 

x²+ y² – 2hx – 2ky = 0

Polar Equation of a Circle

The other form of the equation of the circle is the polar form of the circle, this form of the circle is similar to the parametric form of the circle.

Suppose we take a point on the circumference of the circle (x, y) as (r cosθ, r sinθ) when the line joining the point with the centre of the circle makes an angle θ with the x-axis and the radius of the circle is ‘r’. Also, the centre of the circle is the origin, (0,0) then the equation of the circle is,

x = r cosθ…(i)

and y = r sinθ…(ii)

Now we can easily convert the above equation into the standard form of the circle as,

Squaring and adding eq(i) and eq(ii) we get,

x² + y² = r² cos²θ + r² sin² θ

x² + y² = r²(cos²θ + sin² θ)

Using Trigonometric Identity, cos²θ + sin² θ  =1

x² + y² = r²

(x-0)² + (y-0)² = a² comparing with (x – h)² + (y – k)² = r²  we get 

• Centre = (h, k) = (0, 0)
• Radius = r

Example:

Find the equation of the circle in the polar if the equation of the circle in standard form is x² + y² = 16.

Solution:

Equation of the circle in polar form is,

x = r cosθ 

y = r sinθ

using this in x² + y² = 16

(r cosθ)²  + (r sinθ)² = 16

r²(cos²θ + sin²θ) = 16

r² = 4²

r = 4

Parametric Equation of a Circle

The parametric form of the circle uses (-h + rcosθ, -k + rsinθ) as the general point on the circumference of the circle. The line joining the centre of the circle to the general point makes an angle θ with the x-axis. Thus, the parametric equation of a circle is 

x² + y² + 2hx + 2ky + c = 0

General point on the circumference of the circle is  (-h + rcosθ, -k + rsinθ).

Diameter Form of Circle Equation

The equation of the circle drawn on the straight line joining two points \((x_1, y_1)\) and \((x_2, y_2)\) is

(\(x-x_1\))(\(x-x_2\)) + (\(y-y_1\))(\(y-y_2\)) = 0.

This is known as diameter form of circle.

Example.1 : Find the diameter form of the circle, the coordinates of the end points of whose diameter are (-1, 2) and (4, -3).

Solution : Given coordinates of two ends of diameter of circle i.e. (-1, 2) and (4, -3).

We know that the equation of the circle described on the line segment joining \((x_1, y_1)\) and \((x_2, y_2)\) as a diameter is

(\(x-x_1\))(\(x-x_2\)) + (\(y-y_1\))(\(y-y_2\)) = 0

Here, \(x_1\) = -1, \(x_2\) = 4, \(y_1\) = 2 and \(y_2\) = -3

So, the equation of the required circle is

(x + 1)(x – 4) + (y – 2)(y + 3) = 0

\(x^2 + y^2 – 3x + y – 10\) = 0

Example.2 : Find the diameter form of the circle, drawn on the intercept made by the line 2x + 3y = 6 between the coordinates axes as diameter.

Solution : The line 2x + 3y = 6 meets X and Y axes at A(3, 0) and B(0, 2) respectively

We know that the equation of the circle described on the line segment joining \((x_1, y_1)\) and \((x_2, y_2)\) as a diameter is

(\(x-x_1\))(\(x-x_2\)) + (\(y-y_1\))(\(y-y_2\)) = 0

Taking AB as a diameter, the equation of the required circle is

(x – 3)(x – 0) + (y – 0)(y – 2) = 0

\(x^2 + y^2 – 3x – 2y\) = 0

** Students can also solve the problems by below method

Example 3: Find the equation of a circle when the endpoints of the diameter are (3, 2) and (5, 8).

Solution: 

Given,

End points of the diameter are (3, 2) and (5, 8)

As we know radius is the mid point of the diameter then,

Coordinate of the center are, [(3+5)/2, (2+8)/2] = (4, 5)

Length of radius = √[(4-3)² + (5-2)²] = √(1+9) = √(10)

Equation of the circle is,

(x-h)² + (y-k)² = r²

where, (h, k) is centre and r is the radius,

Now, 

(x-4)² + (y-5)² = [√(10)]²

The required equation of the circle is (x-4)² + (y-5)² = 10 whose end points of the diameter are (3, 2) and (5, 8).

Equation of Tangents and Normal

The Equation of Tangents and Normal are explained below. Let the equation of the circle be

x² + y² + 2gx + 2fy + c = 0

A tangent at point P(x₁, y₁).

Equation of Tangent of Circle

\(\begin{array}{l}xx_{1}^{{}}+y{{y}_{1}}+g(x+{{x}_{1}})+f(y+{{y}_{1}})+c=0\end{array} \)

Tangent having slope ‘m’

y = mx + C

Where,

\(\begin{array}{l}c=\pm \left( \sqrt{{{g}^{2}}+{{f}^{2}}-c} \right)\left( \sqrt{1+{{m}^{2}}} \right)\end{array} \)

\(\begin{array}{l}c=\pm \sqrt[r]{1+{{m}^{2}}}\end{array} \)

Pair of tangents from external point p (x₁, y₁)

T² = ss₁

Where,

\(\begin{array}{l}T\equiv x{{x}_{1}}+y{{y}_{1}}+g(x+{{x}_{1}})+f(y+{{y}_{1}})\end{array} \)

\(\begin{array}{l}S={{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\end{array} \)

and

\(\begin{array}{l}{{S}_{1}}\equiv x{{{}_{1}}^{2}}+{{y}_{1}}^{2}+2g{{x}_{1}}+2f{{y}_{1}}+c=0\end{array} \)

Equation of normal at p(x₁, y₁) to the circle

\(\begin{array}{l}S\equiv {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\end{array} \)

is

\(\begin{array}{l}\frac{x-{{x}_{1}}}{{{x}_{1}}+g}=\frac{y-{{y}_{1}}}{{{y}_{1}}+f}\end{array} \)

Equation of Chord of a Circle

Equation of chord PQ

\(\begin{array}{l}wher{{e}_{\downarrow }}\,T={{S}_{1}}\end{array} \)

\(\begin{array}{l}T\equiv x{{x}_{1}}+y{{y}_{1}}+g(x+{{x}_{1}})+f(y+{{y}_{1}})+c=0\end{array} \)

\(\begin{array}{l}{{S}_{1}}\equiv x_{1}^{2}+y_{1}^{2}+2g{{x}_{1}}+2f{{y}_{1}}+c=0\end{array} \)

Chord of Contact of a Circle

AB is called the chord of contact. The equation of contact is T = 0.

\(\begin{array}{l}x{{x}_{1}}+y{{y}_{1}}+g(x+{{x}_{1}})+f(y+{{y}_{1}})+c=0\end{array} \)

Radical Axis to the Two Circles

Equation of radical axis to the two circles S₁ and S_2,

\(\begin{array}{l}{{S}_{1}}\equiv {{x}^{2}}+{{y}^{2}}+2{{g}_{1}}x+2{{f}_{1}}y+{{c}_{1}}=0\end{array} \)

and

\(\begin{array}{l}{{S}_{2}}={{x}^{2}}+{{y}^{2}}+2{{g}_{2}}x+2{{f}_{2}}y+{{c}_{2}}=0\end{array} \)

The equation of the radical axis is

S₁ – S₂ = 0

Family of Circles

S₁ + λS₂ = 0

Where ‘λ’ is the parameter

Solved Problems Based on Circles

Illustration 1: Find the centre and the radius of the circle

\(\begin{array}{l}3{{x}^{2}}+3{{y}^{2}}-8x-10y+3=0.\end{array} \)

Solution:

We rewrite the given equation as

\(\begin{array}{l}{{x}^{2}}+{{y}^{2}}-\frac{8}{3}x-\frac{10}{3}y+1=0\end{array} \)

\(\begin{array}{l}\Rightarrow g=-\frac{4}{3},f=-\frac{5}{3},c=1.\end{array} \)

Hence, the centre is (4/3, 5/3), and the radius (r) is

\(\begin{array}{l}\sqrt{\frac{16}{9}+\frac{25}{9}-1}=\sqrt{\frac{32}{9}}=\frac{4\sqrt{2}}{3}.\end{array} \)

Illustration 2: Find the equation of the circle with centre (1, 2) and which passes through the point (4, 6).

Solution:

The radius of the circle is

\(\begin{array}{l}\sqrt{{{\left( 4-1 \right)}^{2}}+{{\left( 6-2 \right)}^{2}}}=\sqrt{25}=5.\end{array} \)

Hence, the equation of the circle is

\(\begin{array}{l}{{\left( x-1 \right)}^{2}}+{{\left( y-2 \right)}^{2}}=25\end{array} \)

\(\begin{array}{l}\Rightarrow {{x}^{2}}+{{y}^{2}}-2x-4y=20.\end{array} \)

Illustration 3: Find the equation of the circle whose diameter is the line joining the points (-4, 3) and (12, -1). Find also the length of the intercept made by it on the y-axis.

Solution:

The required equation of the circle is

\(\begin{array}{l}\left( x+4 \right)\left( x-12 \right)+\left( y-3 \right)\left( y+1 \right)=0.\end{array} \)

On the y-axis,

\(\begin{array}{l}x=0\Rightarrow -48+{{y}^{2}}-2y-3=0\end{array} \)

\(\begin{array}{l}\Rightarrow {{y}^{2}}-2y-51=0\Rightarrow y=1\pm \sqrt{52}\end{array} \)

Hence, the length of intercept on the y-axis

\(\begin{array}{l}=2\sqrt{52}=4\sqrt{13}.\end{array} \)

Illustration 4: Find the equation of the circle passing through (1, 1), (2, -1) and (3, 2).

Solution:

Let the equation be

\(\begin{array}{l}{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0.\end{array} \)

Substituting the coordinates of the three given points, we get

\(\begin{array}{l}2g+2f+c=-2,\end{array} \)

\(\begin{array}{l}4g-2f+c=-5,\end{array} \)

\(\begin{array}{l}6g+4f+c=-13.\end{array} \)

Solving the above three equations, we obtain

\(\begin{array}{l}f=-1/2;g=-5/2,c=4.\end{array} \)

Hence, the equation of the circle is

\(\begin{array}{l}{{x}^{2}}+{{y}^{2}}-5x-y+4=0.\end{array} \)

Illustration 5: Find the equation of the circle whose centre is (3, 4) and which touches the line 5x + 12y = 1.

Solution:

Let r be the radius of the circle.

Then,

r = distance of the centre, i.e., point (3, 4) from the line 5x + 12y = 1

\(\begin{array}{l}=\left| \frac{15+48-1}{\sqrt{25+144}} \right|=\frac{62}{13}.\end{array} \)

Hence, the required equation of the circle is

\(\begin{array}{l}{{\left( x-3 \right)}^{2}}+{{\left( y-4 \right)}^{2}}={{\left( \frac{62}{13} \right)}^{2}}\end{array} \)

\(\begin{array}{l}\Rightarrow {{x}^{2}}+{{y}^{2}}-6x-8y+\frac{381}{169}=0.\end{array} \)

Illustration 6: Find the greatest distance of the point P(10, 7) from the circle

\(\begin{array}{l}{{x}^{2}}+{{y}^{2}}-4x-2y-20=0.\end{array} \)

Solution:

Since

\(\begin{array}{l}{{S}_{1}}={{10}^{2}}+{{7}^{2}}-4\times 10-2\times 7-20>0,\end{array} \)

P lies outside the circle.

Join P with the centre C(2, 1) of the given circle.

Suppose PC cuts the circle at A and B, where A is nearer to C.

Then, PB is the greatest distance of P from the circle.

We have, 

\(\begin{array}{l}PC=\sqrt{{{\left( 10-2 \right)}^{2}}+{{\left( 7-1 \right)}^{2}}}=10\end{array} \)

and CB = radius

\(\begin{array}{l}=\sqrt{4+1+20}=5\end{array} \)

Therefore, PB = PC + CB = 10 + 5 = 15.

Illustration 7: A foot of the normal from the point (4, 3) to a circle is (2, 1), and the diameter of the circle has the equation 2x – y = 2. Then, find the equation of the circle.

Solution:

The line joining (4, 3) and (2, 1) is also along a diameter.

So, the centre is the point of intersection of the diameter 2x – y = 2 and the line

\(\begin{array}{l}y-3=\frac{3-1}{4-2}\left( x-4 \right)\end{array} \)

, i.e., x – y – 1 = 0.

Solving these, we get the centre as (1, 0).

Also, the radius = the distance between (1, 0) and (2, 1)

\(\begin{array}{l}=\sqrt{2}\end{array} \)

Hence, the equation of circle is

\(\begin{array}{l}{{\left( x-1 \right)}^{2}}+{{\left( y-0 \right)}^{2}}=2\end{array} \)

or

\(\begin{array}{l}{{x}^{2}}+{{y}^{2}}-2x-1=0.\end{array} \)

Illustration 8: Find the length of the common chord of the circles

\(\begin{array}{l}{{x}^{2}}+{{y}^{2}}+2x+6y=0\end{array} \)

and

\(\begin{array}{l}{{x}^{2}}+{{y}^{2}}-4x-2y-6=0\end{array} \)

Solution:

Equation of common chord is

\(\begin{array}{l}{{S}_{1}}-{{S}_{2}}=0\end{array} \)

or

\(\begin{array}{l}6x+8y+6=0\end{array} \)

or

\(\begin{array}{l}3x+4y+3=0.\end{array} \)

Centre of S₁ is C₁(-1, -3) and its radius is

\(\begin{array}{l}\sqrt{1+9}=\sqrt{10}.\end{array} \)

Distance of C₁ from the common chord

\(\begin{array}{l}=\frac{\left| -3-12+3 \right|}{\sqrt{9+16}}=\frac{12}{5}\end{array} \)

Therefore, the length of common chord

\(\begin{array}{l}=2\sqrt{{{\left( \sqrt{10} \right)}^{2}}-{{\left( \frac{12}{5} \right)}^{2}}}=2\sqrt{10-\frac{144}{25}}=2\sqrt{\frac{106}{25}}=\frac{2\sqrt{106}}{5}\end{array} \)

Illustration 9: A pair of perpendicular straight lines pass through the origin and also through the point of intersection of the curve. The set containing the value of a is

(a) {-2, 2}

(b) {-3, 3}

(c) {-4, 4}

(d) {-5, 5}

Solution:

To make the curve

\(\begin{array}{l}{{x}^{2}}+{{y}^{2}}=4\end{array} \)

homogeneous with respect to line

\(\begin{array}{l}x+y=a,\end{array} \)

we should have

\(\begin{array}{l}{{x}^{2}}+{{y}^{2}}-4{{\left( \frac{x+y}{a} \right)}^{2}}=0\end{array} \)

\(\begin{array}{l}\Rightarrow \,\,\,\,\,{{a}^{2}}\left( {{x}^{2}}+{{y}^{2}} \right)-4\left( {{x}^{2}}+{{y}^{2}}+2xy \right)=0\end{array} \)

\(\begin{array}{l}\Rightarrow \,\,\,\,\,{{x}^{2}}\left( {{a}^{2}}-4 \right)+{{y}^{2}}\left( {{a}^{2}}-4 \right)-8xy=0\end{array} \)

Since this is a pair of perpendicular straight lines, we have

∴ a² – 4 + a² – 4 = 0

⇒ a² – 4 = 0

⇒ a = ±2

Therefore, the required set of a is {-2, 2}.

Hence, the correct answer is (a).

Q1

Illustration 10: Find the equation of the circle that passed through the points (1, 0), (-1, 0), and (0, 1).

Solution:

Let the required circle be 

x² + y² + 2gx + 2fy + c = 0

As, it passes through (1, 0), (-1, 0), and (0, 1). Hence, they will satisfy the equation

By substituting them, we get

1 + 2g + c = 0 …(1)

1 – 2g + c = 0 …(2)

1 + 2f + c = 0 …(3)

From eq(1) and (2), we get

g = 0 and c = -1

Now, putting c = -1 in eq(3), we get

f = 0

Now, substituting the values of g, f and c in the main equation, we get

x² + y² = 1

Illustration 11 :: Find the equation of the image of the circle x² + y² + 16x – 8y + 64 = 0 in the line mirror x = 0.

Solution:

Given equation is,

x² + y² + 16x – 8y = – 64

(x² + 16x) + (y² – 8y) = – 64

On adding and subtracting 64 and 16, we get

(x² + 16x + 64) + (y² – 8y + 16) – 64 – 16 = – 64

(x² + 16x + 64) + (y² – 8y + 16) = – 64 + 64 + 16

(x + 8)² + (y – 4)² = 16

(x – (-8))² + (y – 4)² = 4²

As, the radius of this circle is (-8, 4) and radius = 4.

The image of the circle in the line mirror will have its centre as (8, 4) and radius 4.

So, the equation will be,

(x – 8)² + (y – 4)² = 4²

x² + y² – 16x – 8y + 64 = 0

Practice Questions on Equation of Circle

1. Given the equation of a circle (x-2)² + (y+3)² = 49, determine the center and radius of the circle.
2. Write the equation of a circle with a center at (4, -1) and a radius of 5.
3. Find the equation of the line that is tangent to the circle x² + y² = 16 at the point (4, 0).
4. Convert the following general form of a circle equation to its standard form: x² + y² – 6x + 8y + 9 = 0. Then, identify the center and radius of the circle.
5. Determine if the line y = 2x + 3 intersects the circle (x – 3)² + (y – 2)² = 25. If it does, find the points of intersection.
6. Find the equation of a circle of radius 5 units, whose centre lies on the x-axis and which passes through the point (2, 3).
7. Find the equation of a circle with the centre (h, k) and touching the x-axis.
8. Show that the equation x² + y² – 6x + 4y – 36 = 0 represents a circle. Also, find the centre and radius of the circle.

Frequently Asked Questions (FAQs) on Equation of a Circle

Q1

What is the equation for a circle?

The equation for a circle is given by:
(x-h)²+(y-k)² = a²
Where (h,k) is the center and a is the radius of the circle.

Q2

What are the formulas for circles?

The circumference of a circle is equal to 2 (pi) of radius or pi of diameter.
The area of a circle is equal to the pi of radius-squared.

Q3

What is the equation of a circle when the center is at the origin?

At origin, the value of coordinates is (0,0), therefore, the equation of circle becomes:
(x-0)² + (y-0)² = r²
x² + y² = r²

Q4

If (x-4)²+(y+7)²=9 is the equation of circle, then what is the center of circle?

Given, (x-4)²+(y+7)²=9 is the equation of circle. If we compare this equation with the standard equation we get:
(x-h)²+(y-k)² = a²
h=4 and y = -7
Therefore, (4,-7) is the center of circle

Q5

How do we know if an equation is the equation of circle?

If x and y are squared and the coefficient of x² and y² are same, then it is equation of circle. For example, 3x²+3y² = 12 is a circle’s equation.

Q6

What is Equation of Point Circle?

A point circle is the circle whose radius is zero. The equation of the point circle is,

(x-a)² + (y-b)² = 0

Q7

What are Formulas for Circles?

Various formulas for the circle are,

• Circumference of a Circle = 2πr
• Area of a Circle = πr²

Q8

What is Equation of a Circle when the Centre is at the Origin?

When the centre of the circle is at origin (0,0) its equation is given as, (x -0)² + (y-0)² = a²

x² + y² = a²

Q9

What is Equation of a circle when the endpoints of a diameter are given – A (1,3) and B (7,9)?

Equation of the circle when the endpoints of a diameter are given is (x-4)² + (y-6)² = 18.

Q10

What is the equation of a circle given two points A (2,3) and B (6,7)?

The equation of the circle is (x – 4)² + (y – 5)² = 8.