Question

A circle touches a straight line $lx+my+n=0$ and cuts the circle $x^2+y^2=9$ orthogonally. The locus of centres of such circles is

• A. $(lx+my+n{)}^2=(l^2+m^2\left)\right(x^2+y^2-9)$
• B. $(lx+my-n{)}^2=(l^2+m^2\left)\right(x^2+y^2+9)$
• C. $(lx+my+n{)}^2=(l^2+m^2\left)\right(x^2+y^2+9)$
• D. $(lx+my-n{)}^2=(l^2+m^2\left)\right(x^2+y^2-9)$

Answer 1

Answer: (A)

$(lx+my+n{)}^2=(l^2+m^2\left)\right(x^2+y^2-9)$

Let the general equation circle

$x^2+y^2+2gx+2fy+c=0$

Given, this circle is orthogonal to $x^2+y^2-9=0$

Condition of orthogonality

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

$2g\left(0\right)+2f\left(0\right)=c_1-9$

$c_1=9$

$x^2+y^2+2gx+2fy+9=0$ centre $(-g,-f)$

The, distance of centre from line $=$ radius

$\left[\frac{-l\left(g\right)-m\left(f\right)+n}{\sqrt{l^2+m^2}}\right]=\sqrt{(g^2+f^2-9)}$

${(-l(g)-m(f)+n)}^2=(l^2+m^2)(g^2+f^2-9)$

${(-1)}^2{\left(l\right(g)+m(f)-n)}^2=(l^2+m^2)(g^2+f^2-9)$

${(lx+my-n)}^2=(l^2+m^2)(x^2+y^2-9)$