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If a circle C, whose radius is 3, touches externally the circle, x2+y2+2x−4y−4=0 at the point (2,2), then the length of the intercept cut by this circle C, of the x−axis is equal to |
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Answer» Given: Radius of circle is 3 Radius of other circle = \(\sqrt{g^2+f^2-c}\) = \(\sqrt{(-1)^2+2^2-(-4)}=3\) Now, circle C is touching externally other circle at point (2,2) Therefore, (2,2) is midpoint of two centres of the circles as radius are equal. Let the coordinate of centre of circle C be (p,q). Centre of other circle is (−1,2) ⇒ \(\frac{p-1}{2}\) = and \(\frac{q+2}{2}\) = 2 ⇒ p = 5 and q = 2 Equation of circle C is (x − 5)2 + (y − 2)2 =32 ⇒ x2 + y2 −10x − 4y + 20 = 0 x-intercept = \(2\sqrt{g^2-c}\) Here, g = (−5) and c = 20 x-intercept = \(2\sqrt{(-5)^2-20}\) = 2√5 |
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