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
- A \(a^2+b^2=r^2\)
- B \(a^2-b^2=r^2\)
- C \(a^2-b^2+r^2=0\)
- D \(\mathrm{a}^2+\mathrm{b}^2+\mathrm{r}^2=0\)
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…
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