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