Family of Circles: Definition, Formula, Equation, Examples

Circles are fundamental geometric shapes defined as the set of all points equidistant from a fixed central point. This fixed point is known as the centre, and the constant distance from the centre to any point on the circle is called the radius. Circles have rich mathematical properties and appear extensively in both pure and applied mathematics. When studying circles, mathematicians often investigate collections of circles that share certain common properties or relationships, known as families of circles.

Family of Circles

Let’s explore various methods for determining the family of a circle based on specific conditions.

1. Equation of the family of circles passing through the point of intersection of two given circles S = 0 and S’ = 0 is S + λS’ = 0 where λ is the parameter

2. Equation of the family of circles passing through the point of intersection of a given circle S = 0 and a line L = 0is S + λL = 0 where λ is the parameter.

3. Equation of the family of circles touching the given circle S = 0 and the line L = 0 is S + λL=0

4. The equation of the family of circles passing through the two given points P(x₁, y₁) and Q(x₂, y₂) is

$
\left(\mathrm{x}-\mathrm{x}_1\right)\left(\mathrm{x}-\mathrm{x}_2\right)+\left(\mathrm{y}-\mathrm{y}_1\right)\left(\mathrm{y}-\mathrm{y}_2\right)+\lambda\left|\begin{array}{lll}
x & y & 1 \\
x_1 & y_1 & 1 \\
x_2 & y_2 & 1
\end{array}\right|=0
$

5. The equation of the family of circles that touch $\mathrm{y}-\mathrm{y}_1=\mathrm{m}\left(\mathrm{x}-\mathrm{x}_1\right)$ at $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ for any finite m is $\left(\mathrm{x}-\mathrm{x}_1\right)^2+\left(\mathrm{y}-\mathrm{y}_1\right)^2+\lambda\left\{\left(\mathrm{y}-\mathrm{y}_1\right)-\mathrm{m}\left(\mathrm{x}-\mathrm{x}_1\right)\right\}=0$

And if $m$ is infinite then the family of circles is $\left(x-x_1\right)^2+\left(y-y_1\right)^2+\lambda\left(x-x_1\right)=0$

Recommended Video Based on Family of Circles

Solved Examples Based on Family of Circles

Example 1: If circle $x^2+y^2+4 x+8 y-29=0$ bisects the circumference of the circle $x^2+y^2+2 x+3 y+k=0$, then $k$ is equal to

1) 50
2) $\frac{55}{2}$
3) $-\frac{67}{2}$
4) $-\frac{77}{2}$

Solution
Centre of $2^{\text {nd }}$ circle i.e., $\left(-1,-\frac{3}{2}\right)$, must satisfy equation of common chord.
Equation of common chord is $2 \mathrm{x}+5 \mathrm{y}-29-\mathrm{k}=0$

$
\begin{aligned}
& \Rightarrow \quad-2-\frac{15}{2}-29-\mathrm{k}=0 \\
& \Rightarrow \mathrm{k}=-\frac{77}{2}
\end{aligned}
$
Hence, the answer is the option (4).

Example 2: A circle touches the $y$-axis at the point and passes through the point $(2,0)$. Which of the following lines is not tangent to this circle?
1) $4 x-3 y+17=0$

2) $3 x+4 y-6=0$
3) $4 x+3 y-8=0$
4) $3 x-4 y-24=0$

General Form:
The equation of a circle with centre at $(h, k)$ and radius $r$ is

$
\begin{aligned}
& \Rightarrow(x-h)^2+(y-k)^2=r^2 \\
& \Rightarrow x^2+y^2-2 h x-2 k y+h^2+k^2-r^2=0
\end{aligned}
$
Which is of the form :

$
x^2+y^2+2 g x+2 f y+c=0
$
Equation of family of circle touches at $(0,4)(x-0)^2+(y-4)^2+\lambda x=0$
This family of circle passes through $(2,0)$

$
\begin{aligned}
& 4+16+2 \lambda=0 \Rightarrow \lambda=-10 \\
& x^2+y^2-10 x-8 y+16=0
\end{aligned}$

Center $(5,4)$ and radius $=5$
Now check the option
$\begin{aligned}
& 4 x+3 y-8=0 \\
& \left|\frac{4 \times 5+3 \times 4-8}{5}\right|=\frac{24}{5} \neq 5
\end{aligned}$

Example 3: The centre of the circle passing through the point $(0,1)$ and touching the parabola $y=x^2$ at the point $(2,4)$ is:
1) $\left(\frac{-53}{10}, \frac{16}{5}\right)$
2) $\left(\frac{6}{5}, \frac{53}{10}\right)$
3) $\left(\frac{3}{10}, \frac{16}{5}\right)$
4) $\left(\frac{-16}{5}, \frac{53}{10}\right)$

Solution

$\begin{aligned}
& \left.\frac{d y}{d x}\right|_P=4 \\
& (y-4)=4(x-2) \\
& 4 x-y-4=0
\end{aligned}
$
Circle : $(x-2)^2+(y-4)^2+\lambda(4 x-y-4)=0$
passes through $(0,1)$

$
4+9+\lambda(-5)=0 \Rightarrow \lambda=\frac{13}{5}
$
Circle : $x^2+y^2+x(4 \lambda-4)+y(-\lambda-8)+(20-4 \lambda)=0$
Centre : $\left(2-2 \lambda, \frac{\lambda+8}{2}\right) \equiv\left(\frac{-16}{5}, \frac{53}{10}\right)$

Example 4: The circle passes through the intersection of the circle, $x^2+y^2-6 x=0$ and $x^2+y^2-4 y=0$, having its centre on the line, $2 x-3 y+12=0$, also passes through the point.
11) $(-1,3)$
2) $(-3,6)$
3) $(-3,1)$
4) $(1,-3)$

Solution
Let S be the circle passing through point of intersection of $\mathrm{S}_1 \& \mathrm{~S}_2$

$
\begin{aligned}
& \therefore S=S_1+\lambda S_2=0 \\
& \Rightarrow S:\left(x^2+y^2-6 x\right)+\lambda\left(x^2+y^2-4 y\right)=0 \\
& \Rightarrow S: x^2+y^2-\left(\frac{6}{1+\lambda}\right) x-\left(\frac{4 \lambda}{1+\lambda}\right) y=0
\end{aligned}
$

Centre $\left(\frac{3}{1+\lambda}, \frac{2 \lambda}{1+\lambda}\right)$ lies on
$2 x-3 y+12=0 \Rightarrow \lambda=-3$

$
2 x-3 y+12=0 \Rightarrow \lambda=-3
$

Put in (1)

$
\Rightarrow S: x^2+y^2+3 x-6 y=0
$

Now check options point (–3, 6)

Hence, the answer is option (2).

Example 5: Let $
\mathrm{ABCD}
$ be a square of side of unit length. Let a circle $C_1$ centered at A with unit radius is drawn. Another circle $C_{2}$ which touches $C_1$ and the lines AD and AB are tangent to it, is also drawn. Let a tangent line from the point C to the circle $C_2$ meet the side AB at E. If the length of EB is $\alpha +\sqrt{3}\beta$, where $\alpha,\beta$ are integers, then $\alpha+\beta$ is equal to _______________.

1) 0

2) 1

3) 2

4) 3

Solution

Here $A O+O D=1$ or $(\sqrt{2}+1) r=1$

$
\Rightarrow \quad r=\sqrt{2}-1
$

Equation of circle $(x-r)^2+(y-r)^2=r^2$
Equation of $C E$

$
\begin{aligned}
& y-1=m(x-1) \\
& m x-y+1-m=0
\end{aligned}
$

It is tangent to circle

$
\begin{aligned}
\therefore & \left|\frac{m r-r+1-m}{\sqrt{m^2+1}}\right|=r \\
& \left|\frac{(m-1) r+1-m}{\sqrt{m^2+1}}\right|=r \\
& \frac{(m-1)^2(r-1)^2}{m^2+1}=r^2
\end{aligned}
$

Put $r=\sqrt{2}-1$
On solving $m=2-\sqrt{3}, 2+\sqrt{3}$

Taking a greater slope of $C E$ as

$\begin{aligned}
& 2+\sqrt{3} \\
& y-1=(2+\sqrt{3})(x-1) \\
& \text { Put } y=0 \\
& -1=(2+\sqrt{3})(x-1) \\
& \frac{-1}{2+\sqrt{3}} \times\left(\frac{2-\sqrt{3}}{2-\sqrt{3}}\right)=x-1 \\
& x-1=\sqrt{3}-1 \\
& E B=1-x=1-(\sqrt{3}-1) \\
& E B=2-\sqrt{3}
\end{aligned}$