JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
If a circle \(C,\) whose radius is \(3,\) touches externally the circle, \(x^2 + y^2 + 2x - 4y - 4 = 0\) at the point \((2, 2),\) then the length of the intercept cut by circle \(c,\) on the \(x-\) axis is equal to
- A \(\sqrt 5\)
- B \(2\sqrt 3\)
- C \(3\sqrt 2\)
- D \(2\sqrt 5\)
Answer & Solution
Correct Answer
(D) \(2\sqrt 5\)
Step-by-step Solution
Detailed explanation
Given circle is: \({x^2} + {y^2} + 2x - 4y - 4 = 0\) \(\therefore \) it center is \(\left( { - 1,2} \right)\) and radius is \(3\) units. Let \(A=(x,y)\) be the center of the circle \(C\) \(\therefore \frac{{x - 1}}{2} = 2 \Rightarrow x = 5\) \( \Rightarrow y = 2\) So the center…
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