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in Three-dimensional geometry by (35 points)
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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

closed with the note: I got my answer

1 Answer

+1 vote
by (26.1k points)

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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