Find the centre of a circle passing through points (6, -6), (3, -7) and (3, 3).

Asked by Pragya Singh | 1 year ago |  144

1 Answer

Solution :-

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)+ (y + 7)2 = (x – 3)2 + (y – 3)2

(y + 7)2 = (y – 3)2

y+ 14y + 49 = y– 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 ago

Related Questions

In the determine whether the given quadratic equations have real roots and if so, And the roots 3x2 – 2x + 2 = 0

Class 10 Maths Coordinate Geometry View Answer

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

Class 10 Maths Coordinate Geometry View Answer

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 \( P(\sqrt{2} , \sqrt{2}), Q(- \sqrt{2}, – \sqrt{2)} and\; R (-\sqrt{6} , \sqrt{6} )\)

Class 10 Maths Coordinate Geometry View Answer

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

Class 10 Maths Coordinate Geometry View Answer

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

Class 10 Maths Coordinate Geometry View Answer