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AP EAMCET · Maths · Circle

The condition that the lines joining the origin to the points of intersection of the line \(\frac{x}{a}+\frac{y}{b}=2\) and the circle \((x-a)^2+(y-b)^2=r^2\) are at right angles is

  1. A \(a^2+b^2=r^2\)
  2. B \(a^2-b^2=r^2\)
  3. C \(a^2-b^2+r^2=0\)
  4. D \(\mathrm{a}^2+\mathrm{b}^2+\mathrm{r}^2=0\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(a^2+b^2=r^2\)

Step-by-step Solution

Detailed explanation

\(\because \frac{x}{a}+\frac{y}{b}=2 \Rightarrow \frac{x}{2 a}+\frac{y}{2 b}=1\) \[ \begin{aligned} & \&(x-a)^2+(y-b)^2=r^2 \\ & \Rightarrow x^2+y^2-2 a x-2 b y+a^2+b^2-r^2=0 \end{aligned} \] Homogenizing eqn. (ii), we get…