Let A = (6, -6), B = (3, -7), C = (3, 3) are the points on a circle.

If O is the centre, then OA = OB = OC (radii are equal)

If O = (x, y) then

OA = \( \sqrt{(x – 6)^2 + (y + 6)}^2\)

OB = \( \sqrt{(x – 3)^2 + (y + 7})^2 \)

OC = \( \sqrt{(x-3)^2+(y – 3)^2}\)

Choose: OA = OB, we have

After simplifying above, we get -6x = 2y – 14 ….(1)

Similarly: OB = OC

(x – 3)^{2 }+ (y + 7)^{2} = (x – 3)^{2} + (y – 3)^{2}

(y + 7)^{2} = (y – 3)^{2}

y^{2 }+ 14y + 49 = y^{2 }– 6y + 9

20y =-40

or y = -2

Substituting the value of y in equation (1), we get;

-6x = 2y – 14

-6x = -4 – 14 = -18

x = 3

Hence, the centre of the circle located at point (3,-2).

Answered by Abhisek | 1 year agoIn the determine whether the given quadratic equations have real roots and if so, And the roots 3x^{2} – 2x + 2 = 0

Find the point on x-axis which is equidistant from the points (-2, 5) and (2, -3).

Answer the following questions:-

**(i) **Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square.

**(ii)** Prove that the points A (2, 3), B (-2, 2), C (-1, -2) and D (3, -1) are the vertices of a square ABCD.

**(iii)** Name the type of triangle PQR formed by the point \( P(\sqrt{2} , \sqrt{2}), Q(- \sqrt{2}, – \sqrt{2)} and\; R (-\sqrt{6} , \sqrt{6} )\).

Find a point on the x-axis which is equidistant from the points (7, 6) and (-3, 4).

Find the coordinates of the circumcentre of the triangle whose vertices are (3, 0), (-1, -6) and (4, -1). Also, find its circumradius.