Question

If a circle passes through the point $(3,4)$ and cuts $x^2+y^2=9$ orthogonally, then the locus of its centre is $3x+4y=\lambda$. Then $\lambda =$

• A. $11$
• B. $13$
• C. $17$
• D. $23$

Answer 1

The correct option is C $17$

Two circles are orthogonal, if the angle between them is ${90}^o.$

The condition which represents this is $2g_1{g}_2+2f_1{f}_2=c_1+c_2$ for two circles $x^2+y^2+2g_{1}x+2f_{1}y+c_1=0$ and $x^2+y^2+2g_{2}x+2f_{2}y+c_2=0$

Here, let the unknown circle be $x^2+y^2+2gx+2fy+c=0$

The orthogonality condition thus becomes $0+0=-9+c$, implying $c$ to be $9$.

Since the circle passes through $(3,4)$, we substitute that point in the circle to get $9+16+6g+8f+9=0$

i.e. $6g+8f+34=0$

$3g+4f+17=0$

Since the centre of the circle is $(-g,-f),x=-g$ and $y=-f$

$\Rightarrow$ locus is $-3x-4y+17=0$

i.e. $3x+4y=17$