A circle passes through the points $(2, 3)$ and $(4, 5)$ . If its centre lies on the line, $y - 4 x + 3 = 0$ , radius is equal to

1

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2

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\sqrt{5}

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\sqrt{2}

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Solution

The correct option is D $2$
$W e k n o w t h a t t h e e q u a t i o n o f a c i r c l e i s : (x - p) ² + (y - q) ² = r ², s o : (2 - p) ² + (3 - q) ² = r ² (4 - p) ² + (5 - q) ² = r ² 4 - 4 p + p ² + 9 - 6 q + q ² = r ² 16 - 8 p + p ² + 25 - 10 q + q ² = r ² / * (- 1) 4 - 4 p + p ² + 9 - 6 q + q ² = r ² + - 16 + 8 p - p ² - 25 + 10 q - q ² = r ² 4 p + 4 q = 28, a n d w e k n o w t h a t a c e n t e r l i e s o n t h e l i n e : - 4 p + q = - 3 5 q = 25, q = 25 : 5, q = 5, p = 2 W e w i l l p u t i t i n t o t h e f i r s t e q u a t i o n : (2 - 2) ² + (3 - 5) ² = r ² r ² = 4 r = 2.$

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