A circle is a set of points in a plane which are all the same distance from a single point, referred to as the center. The distance from the center to any point on the circle is known as the radius. This article provides an overview of circles, including the equation of tangents, normals, and chords of contact.

Equation of a Circle

The equation of a circle with center at $(h, k)$ and radius $r$ is given by:

$$(x - h)^2 + (y - k)^2 = r^2$$

1. Standard Equation of a Circle: $$x^2 + y^2 = r^2$$

$$\begin{array}{l}{x}^{2}+{y}^{2}={r}^{2}\end{array}$$

Centre = (0, 0) and

Radius = r

2. Equation of a circle in center-radius form: $(x - h)^2 + (y - k)^2 = r^2$

\(\left(x-h\right)^2 + \left(y-k\right)^2 = r^2\)

The centre of the circle is $(h, k)$ and the radius is $r$.

3. Equation of a Circle in General Form: $$x^2 + y^2 + 2gx + 2fy + c = 0$$

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

Centre = $(-g, -f)$

r2 = g2 + f2 - c

(\sqrt{{{g}^{2}}+{{f}^{2}}-c} = \text{Radius})

(x - x1)^2 + (y - y1)^2 = (x2 - x1)^2 + (y2 - y1)^2

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

5. Equation of circle through three non-collinear points P(x1, y1), Q(x2, y2) and R(x3, y3) is:

$$(x-x_1)^2 + (y-y_1)^2 = (x-x_2)^2 + (y-y_2)^2 = (x-x_3)^2 + (y-y_3)^2$$

(\begin{vmatrix} x^2 + y^2 & x & y & 1 \ x_1^2 + y_1^2 & x_1 & y_1 & 1 \ x_2^2 + y_2^2 & x_2 & y_2 & 1 \ x_3^2 + y_3^2 & x_3 & y_3 & 1 \end{vmatrix} = 0)

Area of circle = πr²

Perimeter = 2πr, where r is the radius.

Equation of a Circle under Different Conditions

(x - a)² + (y - a)² = a²

The equation of the circle touching both axis with centre (a, a) and radius r = a is:

(x - a)^2 + (y - a)^2 = a^2

(\begin{array}{l}{(x - a)^2} + {(y - b)^2} = {a^2}\end{array})

Touches the x-axis only, with center $(\alpha, a)$

(\begin{array}{l}{(x-\alpha)}^2 + {(y-\beta)}^2 = {a}^2\end{array})

The graph of the equation y = β touches the y-axis only at (0, β)

(\begin{array}{l}{(x-a)^2} + {(y-\beta)^2} = {a^2}\end{array})

(x - (α/2))^2 + (y - (β/2))^2 = r^2

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

Parametric Equation of a Circle

Equation of a circle: x^2 + y^2 = r^2

X = r * cos(θ)

Y = r sin θ

Squaring both sides:

x² + y² = r² (cos²θ + sin²θ)

r^2 (cos^2θ + sin^2θ)

x^2 + y^2 = r^2

Position of a point with respect to a circle

The circle is given by the equation x^2 + y^2 + 2gx + 2fy + c = 0 and the point p(x1, y1) is given.

R - Radius

R > cp, {Point lies outside}

CP = R, _on the curve_

cp < R, {x | x ∈ ℝ, x < R}

Parametric Equation of a Circle - Video Lesson

Equation of Tangents and Normals

The Equation of Tangents and Normal is explained below. Let the equation of a circle be:

x^2 + y^2 + 2gx + 2fy + c = 0

A tangent line at point P(x1, y1).

Tangent of a Circle Equation

(\begin{array}{l}xx_{1} + yy_{1} + g(x + x_{1}) + f(y + y_{1}) + c = 0\end{array})

The slope of the tangent is ’m'.

y = mx + b

Where $$c=\pm \left( \sqrt{{{g}^{2}}+{{f}^{2}}-c} \right)\left( \sqrt{1+{{m}^{2}}} \right)$$

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

Pair of tangents from an external point $p (x_1, y_1)$

T2 = ss1

Where (T \equiv x_1x + y_1y + g(x + x_1) + f(y + y_1))

(\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 + 2gx_1 + 2fy_1 + c = 0\end{array})

The equation of the normal at point $(x_1, y_1)$ to the circle $$S \equiv x^2 + y^2 + 2gx + 2fy + c = 0$$ is

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

Chord Equation

Equation of chord PQ:

$$\overline{PQ} = \overline{x_1,y_1} - \overline{x_2,y_2}$$

(\begin{array}{l}where\ T=S_{1}\end{array})

(\begin{array}{l}T = x_1x + y_1y + g(x + x_1) + f(y + y_1) + c = 0\end{array} )

(\begin{array}{l}{S_1} \equiv x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c = 0\end{array})

Chord of Connection

The chord of contact is referred to as AB. The equation of contact is given by T = 0.

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

Radical Axis of the Two Circles

Equation of Radical Axis of the two circles S1 and S2.

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

$$S_2 = x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$$

Equation of Radical Axis is: $$x^2/a^2 + y^2/b^2 = 1$$

S1 - S2 = 0

Family of Circles

S1 + λS2 = 0

Where λ is the parameter

Particular Cases of a Circle

![Particular Cases of Circle]()

Problems 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} )

Given:

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Solution:

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

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

The center is (4/3, 5/3) and the radius (r) is

(\sqrt{\frac{16}{9}+\frac{25}{9}-1} = \frac{4\sqrt{2}}{3})

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

Given:

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Solution:

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The radius of the circle is $$\sqrt{{{\left( 4-1 \right)}^{2}}+{{\left( 6-2 \right)}^{2}}}=\sqrt{25}=5.$$

Hence, the equation of the circle is:

(\left( x-1 \right)^2 + \left( y-2 \right)^2 = 25)

(\begin{array}{l}{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 intercept made by it on the $y$-axis.

Given:

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Solution:

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The equation of a circle is: $(x - h)^2 + (y - k)^2 = r^2$

(\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, $$x=0 \Rightarrow -48 + y^2 - 2y - 3 = 0$$

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

Hence the length of intercept on the y-axis $$=2\sqrt{52}=4\sqrt{13}.$$

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

Given:

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Solution:

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

Substituting the coordinates of the three given points, we obtain.

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

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

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

Solving the above three equations, we obtain:

(\begin{array}{l}f=-\frac{1}{2};g=-\frac{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: The equation of the circle whose centre is (3, 4) and which touches the line 5x + 12y = 1 is (x-3)^2 + (y-4)^2 = 25.

Given: This statement is bold.

Solution: This statement is bold.

The circumference of the circle is equal to $2\pi r$.

r = $\sqrt{(3-0)^2 + (4-0)^2}$ = $\sqrt{25}$

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

Hence the required equation of the circle is $$\left( x-3 \right)^2 + \left( y-4 \right)^2 = \left( \frac{62}{13} \right)^2$$

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

Given:

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Solution:

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P lies outside the circle, since $$S_1 = 10^2 + 7^2 - 4 \times 10 - 2 \times 7 - 20 > 0.$$

Connect point P to the center C(2, 1) of the given circle.

Suppose PC cuts the circle at points A and B, with A being nearer to C.

PB is the greatest distance of P from the circle.

We have, $$PC=\sqrt{{{\left( 10-2 \right)}^{2}}+{{\left( 7-1 \right)}^{2}}}=10$$

and CB = radius $$=\sqrt{4+1+20}=5$$

∴ PB = PC + CB = 10 + 5 = 15

Illustration 7: The equation of the circle with a diameter of 2x – y = 2 and a foot from the point (4, 3) to the circle at (2, 1) is (x - 1)^2 + (y - 2)^2 = 5.

Given:

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Solution:

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

The centre is the point of intersection of the diameter (2x - y = 2) and the line (y - 3 = \frac{3 - 1}{4 - 2}\left(x - 4\right)) i.e., (x - y - 1 = 0)

We get the centre as (1, 0) by solving these.

The radius = the distance between (1, 0) and (2, 1) = $\sqrt{2}$

The equation of a circle can be expressed as either (\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 $\begin{array}{l}{{x}^{2}}+{{y}^{2}}-4x-2y-6=0\end{array}$

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Solution: **This is *bold* text.**

The equation of a common chord is $$S_1 - S_2 = 0$$ or $$6x + 8y + 6 = 0$$ or $$3x + 4y + 3 = 0.$$

The centre of S1 is C1(-1, -3) and its radius is (\sqrt{10}).

Distance of C1 from the common chord = $\frac{12}{5}$

Length of common chord = $\frac{2\sqrt{106}}{5}$

Illustration 9: Two perpendicular straight lines passing through the origin and the point of intersection of the curve with the set containing the value of a.

( \left{-2, 2\right} )

{-3, 3}

{-4, 4}

{-5, 5}

Given:

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Solution:

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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 multiply both sides by (a^2) to obtain: (\begin{array}{l}{{x}^{2}}a^2+{{y}^{2}}a^2=4a^2\end{array} )

(\begin{array}{l}{{x}^{2}} + {{y}^{2}} - 4\left(x + y\right)^2/a^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 these are two perpendicular straight lines, we can conclude that

a² - 4 + a² - 4 = 0

a2 - 4 = 0

a = ±2

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

Therefore, option (a) is the correct answer.

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Frequently Asked Questions

The equation of a Circle with centre (0, 0) and radius, r is: $$x^2 + y^2 = r^2$$

The standard equation of a Circle with centre (0, 0) and radius, r is given by x² + y² = r².

A chord of a circle is a line segment that connects two points on the circle.

A chord is a line segment joining two points on the circle. The diameter is the largest chord.

The standard equation of a Circle with centre $(h, k)$ and radius $r$ is given by: $$(x-h)^2 + (y-k)^2 = r^2$$

The standard equation of a Circle with centre $(h, k)$ and radius, $r$ is given by $(x-h)^2 + (y-k)^2 = r^2$.