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.

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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, $|CP|=r$

Using the distance formula, we have

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

If the centre of the circle is the origin or (0,0) then the equation of the circle becomes$\begin{array}{rl}& (x-0{)}^2+(y-0{)}^2=r^2\\ & \text{i.e.}{x}^2+y^2=r^2\end{array}$

Power of a point wrt Circle

The power of a point $\mathrm{P}(\mathrm{a},\mathrm{b})$ with respect to the circle $S:{\mathrm{x}}^2+{\mathrm{y}}^2+2g\mathrm{g}+2\mathrm{fy}+\mathrm{c}=0$ 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 $\mathrm{P}({\mathrm{x}}_1,{\mathrm{y}}_1)$ with respect to the circle $x^2+y^2+2gx+2fy+c=0$, we use the formula $\mathrm{T}=0$
So the equation of chord of contact is $x{x}_1+y{y}_1+g(x+x_1)+f(y+y_1)+c=0$ mathematician's toolkit.

Recommended Video Based on Power of a Point wrt Circle

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 $2x^2+2y^2+8x-8y+8=0$

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 $2x^2+2y^2+8x-8y+8=0$
$\Rightarrow x^2+y^2+4x-4y+4=0$
length of tangent $=\sqrt{\left(S_1\right)}=\sqrt{4}=2$

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

1) 12

2) 3

3) 16

4) None of these

Solution

We know

$\begin{array}{rl}& PA\cdot PB=S_1\\ & S_1=(2{)}^2+(-1{)}^2+4\times 2+4\times -1+4\\ & PA\cdot PB=13\end{array}$
Hence, the answer is the option 2.

Example 3: Length of a tangent from a point $(-5,-4)$ to the circle $x^2+y^2-4x+2x-10=0$ 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
$x^2+y^2-2(t^2-3t+1)x-2(t^2+2t)y+t=0$, 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) $(\frac{1}{10},-\frac{1}{10})$
2) $(-\frac{1}{10},\frac{1}{10})$
3) $(\frac{1}{10},\frac{1}{10})$
4) $(-\frac{1}{10},-\frac{1}{10})$

Solution
Power:

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

This power is independent of the parameter $t$ if and only if: $2a+2b=0\Rightarrow a=-b$
and $6\mathrm{a}-4\mathrm{b}+1=0$
$\Rightarrow \mathrm{a}=-\frac{1}{10}$ 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) $\mathrm{c}=r({\lambda }^2+{\mu }^2)$
2) $\mathrm{r}=\mathrm{c}({\lambda }^2+{\mu }^2)$
3) ${\mathrm{c}}^2={\mathrm{r}}^2({\lambda }^2+{\mu }^2)$
4) ${\mathrm{r}}^2={\mathrm{c}}^2({\lambda }^2+{\mu }^2)$

Solution

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

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

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

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

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

Hence, the answer is the option (2).