Question

The centre of a circle$S=0$lies on$2x-2y+9=0$and $S=0$cuts orthogonally the circle $x^2+y^2=4.$ Then the circle must pass through the point

• A. $(1,1)$
• B. $(-\frac{1}{2},\frac{1}{2})$
• C. $(5,5)$
• D. $(-4,4)$

Answer 1

Answer: (B), (D)

$(-4,4)$

Step 1:Finding the value of $c$

Let the circle be $S:x^2+y^2+2gx+2fy+c=0..\left(i\right)$

Centre of the circle is $(-g,-f).$

Centre lies on $2x–2y+9=0$

$\Rightarrow -2g+2f+9=0...\left(ii\right)$

It cuts orthogonally the circle $x^2+y^2-4=0$

Here $g_2=0,f_2=0,c_2=-4$

So $2g_1{g}_2+2f_1{f}_2=c_1+c_2$

$$\begin{gathered} \Rightarrow 0+0=c–4 \\ \Rightarrow c=4 \end{gathered}$$

Step 2: Forming the equation of the circle

From $\left(ii\right),2g=2f+9$

Put $c$ and $2g$in equation $\left(i\right)$

We get : $x^2+y^2+(2f+9)x+2fy+4=0$

$\Rightarrow x^2+y^2+9x+4+2f(x+y)=0$

This is of the form $S+\lambda P=0$

Above represents a family of circles which passes through the points of intersection of$S=0$and $P=0$

Step 3: Finding the value of $x$and $y$ i.e the required coordinate

We have to solve $x^2+y^2+9x+4=0..\left(iii\right)$

and $x+y=0.$

$$\begin{gathered} x+y=0 \\ \Rightarrow y=-x \end{gathered}$$

Put $y=-x$ in $\left(iii\right)$

$$\begin{gathered} \Rightarrow x^2+x^2+9x+4=0. \\ \Rightarrow 2x^2+9x+4=0 \\ \Rightarrow 2x^2+8x+x+4=0 \\ \Rightarrow 2x(x+4)+x+4=0 \\ \Rightarrow (2x+1)(x+4)=0 \\ \Rightarrow x=-\frac{1}{2}orx=-4 \end{gathered}$$

When $x=-\frac{1}{2}$, then $y=\frac{1}{2}$

When $x=-4$, then $y=4$

So the circle must pass through the points are $\left(-\frac{1}{2},\frac{1}{2}\right)$ and $(-4,4)$

Hence, option b and d are correct.