Question:
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 this circle C, on the x-axis is equal to :
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 this circle C, on the x-axis is equal to :
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The given circle has centre (-1, 2) and radius 3; for external contact with equal radii the point of contact is the midpoint of the centres.
Updated On: Nov 11, 2026
- \(2\sqrt{5}\)
- \(3\sqrt{2}\)
- \(\sqrt{5}\)
- \(2\sqrt{3}\)
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The Correct Option is A
Solution and Explanation
The given circle: centre \((-1, 2)\), radius \(\sqrt{1+4+4} = 3\).
Circle C has the same radius 3 and touches it externally at (2, 2), so (2, 2) is the midpoint of the two centres: \(\frac{-1+h}{2} = 2,\ \frac{2+k}{2} = 2\), giving centre \((5, 2)\).
Circle C: \((x-5)^2 + (y-2)^2 = 9\).
Put y = 0: \((x-5)^2 = 5\), so \(x = 5 \pm \sqrt5\).
Intercept on the x-axis \(= 2\sqrt5\).
Answer: (1) \(2\sqrt{5}\)
Circle C has the same radius 3 and touches it externally at (2, 2), so (2, 2) is the midpoint of the two centres: \(\frac{-1+h}{2} = 2,\ \frac{2+k}{2} = 2\), giving centre \((5, 2)\).
Circle C: \((x-5)^2 + (y-2)^2 = 9\).
Put y = 0: \((x-5)^2 = 5\), so \(x = 5 \pm \sqrt5\).
Intercept on the x-axis \(= 2\sqrt5\).
Answer: (1) \(2\sqrt{5}\)
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