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 | 1 year agoAn equilateral triangle is inscribed in the parabola y2 = 4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.
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