Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).

Asked by Abhisek | 11 months ago |  77

##### Solution :-

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

We know that the radius of the circle is 5 and its centre lies on the x-axis, k = 0 and r = 5.

So now, the equation of the circle is (x – h)2 + y2 = 25.

It is given that the circle passes through the point (2, 3) so the point will satisfy the equation of the circle.

(2 – h)2+ 32 = 25

(2 – h)2 = 25-9

(2 – h)2 = 16

2 – h = ± $$\sqrt{16}$$ = ± 4

If 2-h = 4, then h = -2

If 2-h = -4, then h = 6

Then, when h = -2, the equation of the circle becomes

(x + 2)2 + y2 = 25

x2 + 12x + 36 + y2 = 25

x2 + y2 + 4x – 21 = 0

When h = 6, the equation of the circle becomes

(x – 6)2 + y2 = 25

x2 -12x + 36 + y2 = 25

x2 + y2 -12x + 11 = 0

The equation of the required circle is x2 + y2 + 4x – 21 = 0 and x2 + y2 -12x + 11 = 0

Answered by Pragya Singh | 11 months ago

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