JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let \(P\) be a moving point on the circle \(x^2 + y^2 - 6x - 8y + 21 = 0\). Then, the maximum distance of \(P\) from the vertex of the parabola \(x^2 + 6x + y + 13 = 0\) is equal to:
- A \(8\)
- B \(10\)
- C \(12\)
- D \(9\)
Answer & Solution
Correct Answer
(C) \(12\)
Step-by-step Solution
Detailed explanation
The equation of the given circle is \(x^2 + y^2 - 6x - 8y + 21 = 0\). The center of the circle is \(C(3, 4)\) and its radius is \(r = \sqrt{3^2 + 4^2 - 21} = \sqrt{4} = 2\). The equation of the given parabola is \(x^2 + 6x + y + 13 = 0\). Rewriting the equation by completing the…
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