Q87 / 180 · COMEDK 2016 · Mathematics

The circles $a x^{2}+a y^{2}+2 g_{1} x+2 f_{1} y+c_{1}=0$ and $b x^{2}+b y^{2}+2 g_{2} x+2 f_{2} y+c_{2}=0$ $(a \neq 0$ and $b \neq 0)$ cut orthogonally if

  • A
    $g_{1} g_{2}+f_{1} f_{2}=c_{1}+c_{2}$
  • B
    $b g_{1} g_{2}+a f_{1} f_{2}=b c_{1}+a c_{2}$
  • Correct answer
    C
    $2(g_{1} g_{2}+f_{1} f_{2})=b c_{1}+a c_{2}$
  • D
    $g_{1} g_{2}+f_{1} f_{2}=a c_{1}+b c_{2}$

Solution

Given equation of circles are $a x^{2}+a y^{2}+2 g_{1} x+2 f_{1} y+c_{1}=0$ and $b x^{2}+b y^{2}+2 g_{2} x+2 f_{2} y+c_{2}=0$ It can be rewritten as, $x^{2}+y^{2}+\frac{2 g_{1}}{a} x+\frac{2 f_{1}}{a} y+\frac{c_{1}}{a}=0$ and $x^{2}+y^{2}+\frac{2 g_{2}}{b} x+\frac{2 f_{2}}{b} y+\frac{c_{2}}{b}=0$ So, centres of circle $\left(-\frac{g_{1}}{a},-\frac{f_{1}}{a}\right)$ and $\left(-\frac{g_{2}}{b},-\frac{f_{2}}{b}\right)$ respectively. We know that, if two circles cut orthogonally, then $\begin{aligned} & 2\left(g_{1} g_{\hat{\mathrm{z}}}+f_{\mathrm{i}} f_{\hat{2}}\right)=c_{\mathrm{i}}+c_{\dot{\mathrm{z}}} \\ \Rightarrow & 2\left(\frac{\mathrm{g}_{1} g_{2}}{a b}+\frac{f_{1} f_{2}}{a b}\right)=\frac{c_{1}}{a}+\frac{c_{2}}{b} \\ & 2\left(g_{1} g_{2}+f_{1} f_{2}\right)=b c_{1}+a c_{2} \end{aligned}$

Hence, the correct answer is option: 'C'.

More Circle Questions