Power of a point wrt Circle

Equation of circle

A circle is the locus of a moving point such that its distance from a fixed point is constant.

The fixed point is called the centre (O) of the circle and the constant distance is called its radius (r)

The equation of a circle with centre at C (h,k) and radius r is (x - h)² + (y - k)² = r²

Let P(x, y) be any point on the circle. Then, by definition, $|C P|=r$

Using the distance formula, we have

$\begin{array}{ll} & \sqrt{(x-h)^2+(y-k)^2}=r \text { i.e. } \quad(x-h)^2+(y-k)^2=r^2\end{array}$

If the centre of the circle is the origin or (0,0) then the equation of the circle becomes

$\begin{aligned} & (x-0)^2+(y-0)^2=r^2 \\ & \text { i.e. } x^2+y^2=r^2\end{aligned}$

Power of a point wrt Circle

The power of a point $\mathrm{P}(\mathrm{a}, \mathrm{b})$ with respect to the circle $S: \mathrm{x}^2+\mathrm{y}^2+2 g \mathrm{~g}+2 \mathrm{fy}+\mathrm{c}=0$ is $\mathrm{S}_1$, where $\mathrm{S}_1: \mathrm{a}^2+\mathrm{b}^2+2 \mathrm{ga}+2 \mathrm{fb}+\mathrm{c}=0$

We know $\mathrm{PA} \cdot \mathrm{PB}=(\mathrm{PT})^2$
Also we know that $\mathrm{PT}=\sqrt{S_1}$
So, $\mathrm{PA} \cdot \mathrm{PB}=\mathrm{PT}^2=\mathrm{S}_1$

Chord of Contact

S is a circle and P(x₁,y₁) be an external point to a circle S. A and B are the points of contact of the tangents drawn from P to circle S. Then the chord AB is called the chord of contact of the circle S drawn from an external point P.

To get the equation of the chord of contact of external point $\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ with respect to the circle $x^2+y^2+2 g x+2 f y+c=0$, we use the formula $\mathrm{T}=0$
So the equation of chord of contact is $x x_1+y y_1+g\left(x+x_1\right)+f\left(y+y_1\right)+c=0$ mathematician's toolkit.

Solved Examples Based on Power of a Point wrt Circle

Example 1: Find the length of the tangent from Point $\mathrm{P}(0,0)$ on the circle $2 x^2+2 y^2+8 x-8 y+8=0$

1) 2

2) $2 \sqrt{2}$

3) 4

4) 8

Solution

The power of a point $\mathrm{P}(\mathrm{a}, \mathrm{b})$ with respect to the circle $S: \mathrm{x}^2+\mathrm{y}^2+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0$ is $\mathrm{S}_1$, where $\mathrm{S}_1: \mathrm{a}^2+\mathrm{b}^2+2 \mathrm{ga}+2 \mathrm{fb}+\mathrm{c}=0$.
$\mathrm{PA} \cdot \mathrm{PB}=(\mathrm{PT})^2=\mathrm{S}_1$
From above concept
length of tangent $=\sqrt{\left(S_1\right)}$
Remember: Factor of $x^2$ is 1
Given $2 x^2+2 y^2+8 x-8 y+8=0$
$\Rightarrow x^2+y^2+4 x-4 y+4=0$
length of tangent $=\sqrt{\left(S_1\right)}=\sqrt{4}=2$

Example 2: A line from point $\mathrm{P}(2,-1)$ to the circle $x^2+y^2+4 x+4 y+4=0$ intersects it at A and B, then the value of $P A \cdot P B$ is

1) 12

2) 3

3) 16

4) None of these

Solution

We know

$\begin{aligned}
& P A \cdot P B=S_1 \\
& S_1=(2)^2+(-1)^2+4 \times 2+4 \times-1+4 \\
& P A \cdot P B=13
\end{aligned}$
Hence, the answer is the option 2.

Example 3: Length of a tangent from a point $(-5,-4)$ to the circle $x^2+y^2-4 x+2 x-10=0$ is

1) 7
2) $\sqrt{48}$
3) 9
14) $\sqrt{43}$

Solution
Length of tangent $=\sqrt{(-5)^2+(-4)^2-4(-5)+2(-4)-10}=\sqrt{43}$
Hence, the answer is the option 4.

Example 4: A variable circle C has the equation
$x^2+y^2-2\left(t^2-3 t+1\right) x-2\left(t^2+2 t\right) y+t=0$, where $t$ is a parameter.
If the power of point $P(a, b)$ w.r.t. the circle $C$ is constant then the ordered pair $(a, b)$ is:

1) $\left(\frac{1}{10},-\frac{1}{10}\right)$
2) $\left(-\frac{1}{10}, \frac{1}{10}\right)$
3) $\left(\frac{1}{10}, \frac{1}{10}\right)$
4) $\left(-\frac{1}{10},-\frac{1}{10}\right)$

Solution
Power:

$\begin{aligned}
P & =a^2+b^2-2\left(t^2-3 t+1\right) a-2\left(t^2+2 t\right) b+t \\
& =-(2 a+2 b) t^2+(6 a-4 b+1) t+a^2+b^2-2 a
\end{aligned}$

This power is independent of the parameter $t$ if and only if: $2 a+2 b=0 \quad \Rightarrow \quad a=-b$
and $6 \mathrm{a}-4 \mathrm{~b}+1=0$
$\Rightarrow \quad \mathrm{a}=-\frac{1}{10}$ and $\mathrm{b}=\frac{1}{10}$
Hence, the answer is the option (2).

Example 5: Polar of origin $(0,0)$ w.r.t. the circle $\mathrm{x}^2+\mathrm{y}^2+2 \lambda \mathrm{x}+2 \mu \mathrm{y}+\mathrm{c}=0$ touches the circle $\mathrm{x}^2+\mathrm{y}^2=\mathrm{r}^2$, if

1) $\mathrm{c}=r\left(\lambda^2+\mu^2\right)$
2) $\mathrm{r}=\mathrm{c}\left(\lambda^2+\mu^2\right)$
3) $\mathrm{c}^2=\mathrm{r}^2\left(\lambda^2+\mu^2\right)$
4) $\mathrm{r}^2=\mathrm{c}^2\left(\lambda^2+\mu^2\right)$

Solution

$x^2+y^2+21 x+2 m y+c=0$

Polar of $(0,0)$ is $x .0+0 . y+\lambda(x+0)+\mu(y+0)+c=0$

$\lambda x+\mu y+c=0 \quad \ldots(1)$

(1) will touch the circle $x^2+y^2=r^2$ if the distance of origin from (1) $=r$

$\frac{|0+0 \times \mathrm{c}|}{\sqrt{\mathrm{I}^2+\mathrm{m}^2}}=\mathrm{r} \quad \mathrm{p} \quad \mathrm{c}^2=\mathrm{r}^2\left(\mathrm{I}^2+\mathrm{m}^2\right)$

Hence, the answer is the option (2).