Q118 / 180 · COMEDK 2023 (Evening Shift) · Mathematics
The centre of the circle passing through $(0,0)$ and $(1,0)$ and touching the circle $x^2+y^2=9$ is
- A$\left(\dfrac{1}{2}, \dfrac{3}{2}\right)$
- B$\left(\dfrac{3}{2}, \dfrac{1}{2}\right)$
- Correct answerC$\left(\dfrac{1}{2},-\sqrt{2}\right)$
- D$\left(\dfrac{1}{2}, \dfrac{1}{2}\right)$
Solution
Centre lies on perpendicular bisector of $(0,0)$ and $(1,0)$, which is $x = \dfrac{1}{2}$. So $h = \dfrac{1}{2}$.
$r^2 = h^2 + k^2 = \dfrac{1}{4} + k^2$
Distance between centres $C_1 = \left(\dfrac{1}{2}, k\right)$ and $C_2 = (0,0)$:
$d = \sqrt{\dfrac{1}{4} + k^2}$
For the circles to touch internally: $d = R - r$
$\sqrt{\dfrac{1}{4}+k^2} = 3 - \sqrt{\dfrac{1}{4}+k^2}$
$2\sqrt{\dfrac{1}{4}+k^2} = 3 \Rightarrow \sqrt{\dfrac{1}{4}+k^2} = \dfrac{3}{2}$
$\dfrac{1}{4} + k^2 = \dfrac{9}{4} \Rightarrow k^2 = 2 \Rightarrow k = \pm\sqrt{2}$
Centre $= \left(\dfrac{1}{2}, -\sqrt{2}\right)$
$r^2 = h^2 + k^2 = \dfrac{1}{4} + k^2$
Distance between centres $C_1 = \left(\dfrac{1}{2}, k\right)$ and $C_2 = (0,0)$:
$d = \sqrt{\dfrac{1}{4} + k^2}$
For the circles to touch internally: $d = R - r$
$\sqrt{\dfrac{1}{4}+k^2} = 3 - \sqrt{\dfrac{1}{4}+k^2}$
$2\sqrt{\dfrac{1}{4}+k^2} = 3 \Rightarrow \sqrt{\dfrac{1}{4}+k^2} = \dfrac{3}{2}$
$\dfrac{1}{4} + k^2 = \dfrac{9}{4} \Rightarrow k^2 = 2 \Rightarrow k = \pm\sqrt{2}$
Centre $= \left(\dfrac{1}{2}, -\sqrt{2}\right)$
Hence, the correct answer is option: 'C'.