Question

0. Find the coordinates of the centre of the
circle passing through the points P(6,- 6),
Q(3, - 7) and R(3, 3).​

Answer 1

$\large\underline{\sf{Given- }}$

• A circle passing through the points P(6,- 6), Q(3, - 7) and R(3, 3).

$\large\underline{\sf{To\:Find - }}$

• The coordinates of center of Circle.

$\begin{gathered}\Large{\sf{{\underline{Formula \: Used - }}}}  \end{gathered}$

Distance Formula :-

Let us assume a line segment joining the points A and B, then distance between A and B is given by

$\sf \:AB = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }$

$\sf \: where \: coordinates \: of \: A \: and \: B \: are \: (x_1,y_1) \: and \: (x_2,y_2)$

$\large\underline{\sf{Solution-}}$

Let suppose that coordinates of center be O(x, y).

Since, O is center of circle and coordinates P, Q and R lies on it.

$\sf \: :\implies\: \: OP = OQ = OR = radius \: of \: circle$

Now,

Consider,

$\rm :\longmapsto\:OR \: = \: OQ$

On squaring both sides, we have

$\rm :\longmapsto\: {OR}^{2} = {OQ}^{2}$

$\rm :\longmapsto\: \cancel{(x - 3)}^{2} + {(y - 3)}^{2} = \cancel{(x - 3)}^{2} + {(y + 7)}^{2}$

$\rm :\longmapsto\: \cancel{y}^{2} + 9 - 6y = \cancel{y}^{2} + 49 + 14y$

$\rm :\longmapsto\:14y + 6y = 9 - 49$

$\rm :\longmapsto\:20y = - 40$

$\bf\implies \:y \: = \: - \: 2 - - (1)$

Again,

Consider,

$\rm :\longmapsto\:OP = OR$

On squaring both sides, we have

$\rm :\longmapsto\: {OP}^{2} = {OR}^{2}$

$\rm :\longmapsto\: {(x - 6)}^{2} + {(y + 6)}^{2} = {(x - 3)}^{2} + {(y - 3)}^{2}$

On substituting the value of 'y = - 2', we have

$\rm :\longmapsto\: {(x - 6)}^{2} + {( - 2 + 6)}^{2} = {(x - 3)}^{2} + {( - - 3)}^{2}$

$\rm :\longmapsto\: \cancel{x}^{2} + 36 - 12x + 16 = \cancel{x}^{2} + 9 - 6x + 25$

$\rm :\longmapsto\:52 - 12x = 34 - 6x$

$\rm :\longmapsto\: - 12x + 6x = 34 - 52$

$\rm :\longmapsto\: - 6x = - 18$

$\bf\implies \:x \: = \: 3 - - (2)$

Hence, Coordinates of center of circle is (3, - 2).

Additional Information :-

Section Formula :-

Section Formula is used to find the co ordinates of the point(Q) Which divides the line segment joining the points (B) and (C) internally,

${\underline{\boxed{\sf{\quad \Big(x, y \Big) = \Bigg(\dfrac{mx_2 + nx_1}{m + n} , \: \dfrac{my_2 + ny_1}{m + n}\Bigg) \quad}}}}$

Midpoint Formula :-

Mid Point formula is used to find the Mid points on any line segment.

${\underline{\boxed{\sf{\quad \bigg(\dfrac{x_1 + x_2}{2} \; ,\; \dfrac{y_1 + y_2}{2} \bigg) \quad}}}}$

Answer 2

Answer:

coordinates of the centre of the circle are (3, - 2)