Question

The locus of the centre of the circle, which cuts the circle $x^2+y^2-20x+4=0$ orthogonally and touches the line $x=2$, is

• A. $x^2=16y$
• B. $y^2=4x$
• C. $y^2=16x$
• D. $x^2=4y$

Answer 1

The correct option is C

$y^2=16x$

Explanation for the correct option

Step 1: Drive an equation from the condition that both the circles are orthogonal.

Assume that, the circle which cuts the circle $x^2+y^2-20x+4=0$ orthogonally and touches the line $x=2$, is $x^2+y^2+2gx+2fy+c=0$.

So, the centre and radius of the second circle is $\left(-g,-f\right)$ and $\sqrt{g^2+f^2-c}$ respectively.

Since the condition for two circles to be orthogonal is $2gg\text{'}+2ff\text{'}=c+c\text{'}$.

So, $-20g=c+4...\left[1\right]$

Step 2: Drive an equation from the condition that the required circle touches the given line.

Since, the second circle touches the line $x=2$ that is, $x-2=0$.

The shortest distance $D$ between a line in the form of $Ax+By+C=0$ and a point $(x_1,y_1)$ is given by: $D=\frac{\left|A{x}_1+B{y}_1+C\right|}{\sqrt{A^2+B^2}}$.

Since the shortest distance between the given line and the centre of the given circle is the radius of the circle as the line touches the circle.

So,

$$\begin{gathered} \frac{\left|-g+0-2\right|}{\sqrt{0^2+1^2}}=\sqrt{g^2+f^2-c} \\ \Rightarrow \frac{\left|-g-2\right|}{1}=\sqrt{g^2+f^2-c} \\ \Rightarrow g+2=\sqrt{g^2+f^2-c} \\ \Rightarrow {\left(g+2\right)}^2=g^2+f^2-c \\ \Rightarrow g^2+4+4g=g^2+f^2-c \\ \Rightarrow 4+4g=f^2-c \\ \Rightarrow 4+c=f^2-4g...\left[2\right] \end{gathered}$$

Step 3: Find the locus of the centre of the required circle.

From equation $\left[1\right]$ and equation $\left[2\right]$, we get.

$$\begin{gathered} -20g=f^2-4g \\ \Rightarrow f^2+16g=0 \\ \Rightarrow {(-f)}^2-16(-g)=0 \end{gathered}$$

Since the centre of the second circle is $\left(-g,-f\right)$.

Replace $-g$ with $x$ and $-f$ with $y$.

Therefore, the locus of the centre of the circle, which cuts the circle $x^2+y^2-20x+4=0$ orthogonally and touches the line $x=2$, is $y^2=16x$.

Hence, option C is the correct answer.