The circle passing through the point and touching the -axis at (0, 2) also passes through the point. [2011]

Question

The circle passing through the point $(- 1, 0)$ and touching the $y$ -axis at (0, 2) also passes through the point. [2011]

Smart Achievers · Accepted Answer

(4)

Let centre of the circle be $(h, 2)$ , then radius = $h$

$∴$ Equation of circle becomes ${(x - h)}^{2} + {(y - 2)}^{2} = h^{2}$

Since the circle passes through $(- 1, 0)$

$∴$ ${(- 1 - h)}^{2} + 4 = h^{2} \Rightarrow h = \frac{- 5}{2}$

$∴$ $Centre (- \frac{5}{2}, 2) and r = \frac{5}{2}$

$Now, distance of centre from (- 4, 0) is \frac{5}{2}$

$∴$ $Point (- 4, 0) lies on the circle.$

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