Let us consider the equation of the required circle be (x – h)^{2}+ (y – k)^{2} = r^{2}

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} + y^{2} = 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}+ 3^{2} = 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} + y^{2} = 25

x^{2} + 12x + 36 + y^{2} = 25

x^{2} + y^{2} + 4x – 21 = 0

When h = 6, the equation of the circle becomes

(x – 6)^{2} + y^{2} = 25

x^{2} -12x + 36 + y^{2} = 25

x^{2} + y^{2} -12x + 11 = 0

The equation of the required circle is x^{2} + y^{2} + 4x – 21 = 0 and x^{2} + y^{2} -12x + 11 = 0

An equilateral triangle is inscribed in the parabola y^{2} = 4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

A man running a racecourse notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man.

Find the area of the triangle formed by the lines joining the vertex of the parabola x^{2} = 12y to the ends of its latus rectum.

A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.

An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.