Q. 125.0( 1 Vote )

Find the

Answer :


Let the equation of the circle be (x – h)2 + (y – k)2 = r2

Since the circle passes through, (2, 3) and its radius is 5

(2 – h)2 + (3 – k)2 = 25

or 4 – 4h + h2 + 9 – 6k + k2 = 25

or – 12 - 4h + h2 – 6k + k2 = 0 …..(i)

Also since the centre lies on the x-axis, we have

K = 0 …..(ii)

Putting k = 0 in equation (i) we have

-12 – 4h + h2 = 0

or h2 – 4h – 12 = 0

or h2 – 6h + 12 – 12 = 0

or h(h – 6) + 2(h – 6) = 0

or (h – 6)(h + 2) = 0

or h = 6 or h = -2

Hence, the required equations of the circle are

a) For h = 6

(x – 6)2 + (y – 0)2 = (5)2

or (x2 - 12x + 36 + y2 = 25

or x2 + 4x + 4 + y2 = 25

or x2 + y2 + 4x – 21 = 0.

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