ExamBro
ExamBro
AP EAMCET · Maths · Circle

Circles are drawn through the point \((2,0)\) to cut intercepts of length 5 units on the X -axis. If their centre lie in the first quadrant, then their equation is

  1. A \(3 \mathrm{x}^2+3 \mathrm{y}^2-27 \mathrm{x}-2 \mathrm{ky}+42=0, \mathrm{k} \in \mathbb{R}^{+}\)
  2. B \(\mathrm{x}^2+\mathrm{y}^2-2 \mathrm{kx}-9 \mathrm{y}+14=0, \mathrm{k} \in \mathbb{R}^{+}\)
  3. C \(\mathrm{x}^2+\mathrm{y}^2-9 \mathrm{x}-2 \mathrm{ky}+14=0, \mathrm{k} \in \mathbb{R}^{+}\)
  4. D \(x^2+y^2-9 x-2 k y-42=0, k \in \mathbb{R}^{+}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\mathrm{x}^2+\mathrm{y}^2-9 \mathrm{x}-2 \mathrm{ky}+14=0, \mathrm{k} \in \mathbb{R}^{+}\)

Step-by-step Solution

Detailed explanation

Let the center of the circle be \((h,k)\) and radius \(r\). Given \(h>0, k>0\). Passes through \((2,0)\): \( (2-h)^2 + (0-k)^2 = r^2 \implies (2-h)^2 + k^2 = r^2 \). X-intercept length is 5: \( 2\sqrt{r^2-k^2} = 5 \implies r^2-k^2 = \frac{25}{4} \). Substitute \(r^2\) from the…