Question

From the point $P(16,7)$, tangents $PQ$ and $PR$ are drawn to circle $x^2+y^2-2x-4y-20=0$.

If $C$ is the centre of the circle, then the area of equilateral $PQCR$ is

• A. $450$ sq units
• B. $15$ sq units
• C. $50$ sq units
• D. $75$ sq units

Answer 1

The correct option is D

$75$ sq units

Explanation for the correct option:

Step 1. Find the area of $PQCR$:

Given: the equation of the circle is

$x^2+y^2-2x-4y-20=0$

$\Rightarrow$ ${(x-1)}^2+{(y-2)}^2=20+1^2+2^2$

$\left[\mathbf{\because }{\left(x+g\right)}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}{\left(y+f\right)}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}{\mathit{g}}^{\mathbf{2}}\mathbf{+}{\mathit{f}}^{\mathbf{2}}\mathbf{-}\mathit{c}\right]$

So, radius of the circle $r=\sqrt{g^2+f^2-c}$

$=\sqrt{1+4+20}$

$=\sqrt{25}$

$=5$

Given the coordinates of external point $P=(16,7)$

and the coordinates of the center C $=(1,2)$

Step 2. Find the value of $PC$:

$P{C}^2={(16-1)}^2+{(7-2)}^2$ [$\because$Distance formula $\mathbf{=}\mathbf{}{\mathbf{(}{\mathbf{x}}_{\mathbf{2}}\mathbf{-}{\mathbf{x}}_{\mathbf{1}}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{(}{\mathbf{y}}_{\mathbf{2}}\mathbf{-}{\mathbf{y}}_{\mathbf{1}}\mathbf{)}}^{\mathbf{2}}$]

$={15}^2+5^2$

$=225+25$

$=250$

$\because$ $P{R}^2=P{Q}^2$ [$\because$Tangents from an external point to the circle are equal]

$=P{C}^2-r^2$ [using Pythagoras theorem]

$=250-25$

$=225$

$\Rightarrow$ $PR=PQ$

$=15$

Step 3. Area of $△PCQ$ or $△PCR$ $\mathbf{=}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\times }\mathit{b}\mathbf{\times }\mathit{h}$

$=\frac{1}{2}\times 15\times 5$

$=\frac{75}{2}$ sq. units

$\therefore$Area of $PQCR$ $=2\times$Area of $△PCQ$ or $△PCR$

$=2\times \frac{75}{2}$

$=\mathbf{}\mathbf{75}$ sq. units.

Hence, Option ‘D' is Correct.