Answer :

Let the equation of the circle be (x – h)

^{2}+ (y – k)

^{2}= r

^{2}

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

(2 – h)^{2} + (3 – k)^{2} = 25

or 4 – 4h + h^{2} + 9 – 6k + k^{2} = 25

or – 12 - 4h + h^{2} – 6k + k^{2} = 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 + h^{2} = 0

or h^{2} – 4h – 12 = 0

or h^{2} – 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 (x^{2} - 12x + 36 + y^{2} = 25

or x^{2} + 4x + 4 + y^{2} = 25

or x^{2} + y^{2} + 4x – 21 = 0.

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