Let us consider the equation of the required circle be

(x – h)^{2}+ (y – k)^{2} = r^{2}

We know that the circle passes through points (4,1) and (6,5)

So,

(4 – h)^{2 }+ (1 – k)^{2} = r^{2} ……………..(1)

(6– h)^{2}+ (5 – k)^{2} = r^{2} ………………(2)

Since, the centre (h, k) of the circle lies on line 4x + y = 16,

4h + k =16………………… (3)

From the equation (1) and (2), we obtain

(4 – h)^{2}+ (1 – k)^{2} =(6 – h)^{2} + (5 – k)^{2}

16 – 8h + h^{2} +1 -2k +k^{2} = 36 -12h +h^{2}+15 – 10k + k^{2}

16 – 8h +1 -2k + 12h -25 -10k

4h +8k = 44

h + 2k =11……………. (4)

On solving equations (3) and (4), we obtain h=3 and k= 4.

On substituting the values of h and k in equation (1), we obtain

(4 – 3)^{2}+ (1 – 4)^{2} = r^{2}

(1)^{2} + (-3)^{2} = r^{2}

1+9 = r^{2}

r =\( \sqrt{10}\)

so now, (x – 3)^{2 }+ (y – 4)^{2} = \( (\sqrt{10})^2\)

x^{2} – 6x + 9 + y^{2} – 8y + 16 =10

x^{2} + y^{2} – 6x – 8y + 15 = 0

The equation of the required circle is x^{2} + y^{2} – 6x – 8y + 15 = 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.