The circle passing through (1, – 2) and touching the axis of x at (3, 0), also passes through the point

Engineering

Mathematics

Intercept from a Circle

Question

(– 5, 2)

(5, – 2)

(2, – 5)

(– 2, 5)

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$\sqrt{{(3 - 1)}^{2} + {(y + 2)}^{2}} = | y | = radius$

$\Rightarrow$ 4 + (y + 2)² = y²

$\Rightarrow$ 4y = – 8

$\Rightarrow$ y = – 2

$∴$ Centre (3, – 2) and r = 2

$∴$ (x – 3)² + (y + 2)² = 4