Question

Find the co-ordinates of the circumcenter and radius of circumcircle of PQR if P (5, –1),Q

(–3, 3) and R (–2, 6)​

Answer 1

Step-by-step explanation:

Note :- Refer to above attachment.

As we know that According to properties of Circumcentre(S) distance between S and all the vertex's are same.

$\sf\implies$

SP = SQ = SR

Let's equate SP = SQ

Distance Formula:-

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

SP = SQ

First finding for 'SP' :-

$\sf x_2 = x$

$\sf x_1 = 5$

$\sf y_2 = y$

$\sf y_1 = -1$

SP =

$\sf\sqrt{(x - 5)^{2} + (y + 1)^{2}}$

$\sf\sqrt{x^{2} - 10x + 25 + y^{2} + 2y + 1}$

SP = $\sf\sqrt{x^{2} + y^{2} - 10x + 2y + 26}$

Now for SQ :-

$\sf x_2 = x$

$\sf x_1 = - 3$

$\sf y_2 = y$

$\sf y_1 = 3$

SQ =

$\sf\sqrt{(x + 3)^{2} + (y - 3)^{2}}$

$\sf\sqrt{x^{2} + 6x + 9 + y^{2} - 6y + 9}$

SQ = $\sf\sqrt{x^{2} + y^{2} + 6x - 6y + 18}$

Equating both SP and SQ:-

$\sf\sqrt{x^{2} + y^{2} - 10x + 2y + 26}$ = $\sf\sqrt{x^{2} + y^{2} + 6x - 6y + 18}$

Squaring on both sides :-

$\sf x^{2} + y^{2} - 10x + 2y + 26 = x^{2} + y^{2} + 6x - 6y + 18$

on both sides x^2 and y^2 cancelled:-

-10x - 6x + 2y + 6y = 18 - 26

-16x + 8y = -8

-8(2x - y) = -8

2x - y = 1 (let it be equation(1))

Now ,

SQ = SR

First we find for SQ , in above we already finded so , SQ =

$\sf\sqrt{x^{2} + y^{2} + 6x - 6y + 18}$

for SR =

$\sf x_2 = x$

$\sf x_1 = - 2$

$\sf y_2 = y$

$\sf y_1 = 6$

$\sf\sqrt{(x + 2)^{2} + (y - 6)^{2}}$

$\sf\sqrt{x^{2} + 4x + 4 + y^{2} - 12y + 36}$

SR = $\sf\sqrt{x^{2} + y^{2} + 4x - 12y + 40}$

Equating SQ = SR and squaring on both sides :-

$\sf x^{2} + y^{2} + 6x - 6y + 18 = x^{2} + y^{2} + 4x - 12y + 40$

6x - 4x - 6y + 12y = 40 - 18

2x + 6y = 22(Let it be equation(2))

Equation 2 - Equation 1 we get :-

y = 3

On substituting in equation (1) we get x = 2.

S = (x , y) = (2 , 3)

Circum radius = SR = substituting values of x and y in before value :-

SR = $\sf\sqrt{2^{2} + 3^{2} + 4(2) - 12(3) + 40}$

= $\sf\sqrt{25}$

= 5

$\sf\therefore$ Circum radius = 5units

Co - ordinates of S(Circumcentre) = (2 , 3).