Power of a point wrt Circle

The concept of the "power of a point" relative to a circle is a fascinating and useful idea in geometry. It provides a relationship between a point and a circle that can be used in various geometric proofs and constructions. This concept extends beyond basic circle properties, offering deeper insights into the interactions between points and circles.

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).