Q202 / 270 · NDA 2019 (Phase 1) · Mathematics
A circle is drawn on the chord of a circle $x^2+y^2=a^2$ as diameter. The chord lies on the line $x+y=a$. What is the equation of the circle?
- A$x^2+y^2-a x-a y+a^2=0$
- Correct answerB$x^2+y^2-a x-a y=0$
- C$x^2+y^2+a x+a y=0$
- D$x^2+y^2+a x+a y-2 a^2=0$
Solution
Given, equation of circle $\Rightarrow x^2+y^2=a^2$...(1)
Equation of chord $\Rightarrow x+y=a$...(2)
$\begin{aligned} & \text { (1) } \Rightarrow x^2+(a-x)^2=a^2 \\ & \Rightarrow x^2+a^2+x^2-2 a x=a^2 \\ & \Rightarrow 2 x^2=2 a x \\ & \Rightarrow x=0, a\end{aligned}$
When, $x=0, y=a$ and when $x=a, y=0$.
$\therefore$ Points of intersection are $(0, a)$ and $(a, 0)$
$\therefore$ Equation of circle with chord as diameter is
$\begin{aligned} & (x-0)(x-a)+(y-a)(y-0)=0 \\ & \Rightarrow x(x-a)+y(y-a)=0 \\ & \Rightarrow x^2-a x+y^2-a y=0 \\ & \Rightarrow x^2+y^2-a x-a y=0\end{aligned}$
Hence, the correct answer is option: 'B'.