If the circles $(x+a)^2+(y+b)^2=a^2$ and $(x+c)^2+(y+d)^2=d^2$ cut orthogonally, then $b(b-2 d)=$

Question

Aicharya · Accepted Answer

Given equation of circles are $\begin{aligned} (x+a)^2+(y+b)^2 & =a^2 \ \Rightarrow \quad x^2+y^2+2 a x+2 b y+b^2 & =0 \quad \ldots (i) \ (x+c)^2+(y+d)^2 & =d^2 \ \Rightarrow \quad x^2+y^2+2 c x+2 d y+c^2 & =0 \quad \ldots (ii) \end{aligned}$ From Eq. (i) $g_1=a ; f_1=b, c_1=b^2$ From Eq. (ii) $g_2=c ; f_2=d ; c_2=c^2$ If circles (i) and (ii) are orthogonal, $\begin{array}{rlrl} \Rightarrow & & 2\left[g_1 g_2+f_1 f_2\right] & =c_1+c_2 \ \Rightarrow & 2 & 2(a c+b d) & =b^2+c^2 \ & \Rightarrow & 2 a c+2 b d & =b^2+c^2 \ & \Rightarrow & 2 a c-c^2 & =b^2-2 b d \ & \Rightarrow & c(2 a-c) & =b(b-2 d) \ & \therefore & b(b-2 d) & =c(2 a-c) \end{array}$ $\therefore$ Hence, answer is (b).

If the circles \((x+a)^2+(y+b)^2=a^2\) and \((x+c)^2+(y+d)^2=d^2\) cut orthogonally, then \(b(b-2 d)=\)

Solution