Question

in triangle pqr , if p(5,-1) , q(-3,3) ,r(-2,6) are vertices then find the coordinates of the circumcentre and radius of the circumcircle​

Answer 1

In triangle pqr with vertices p(5,-1) , q(-3,3) ,r(-2,6), the coordinates of the circumcentre is o(3,2) and radius of the circumcircle​ is $\sqrt{13}$

Step-by-step explanation:

Given:

In triangle pqr we have p(5,-1) , q(-3,3) ,r(-2,6).

In the figure given below we have, let say o(l,m) be the centre then

oq = or (radius of the circle)

$\sqrt{(l-(-3))^{2}+(m-3)^{2}  }$ = $\sqrt{(l-(-2))^{2}+(m-6)^{2}  }$

Cancel square roots from both the sides

Then we have,

$(l+3)^{2} +(m-3)^{2}$ = $(l+2)^{2} +(m-6)^{2}$

$l^{2}+6l+9+m^{2}-6m+  9$ = $l^{2}+4l+4+m^{2}-12m+36$

Now cancelling on both the sides, we get

$6l-6m+18$ = $4l-12m+40$

$6l-6m-4l+12m=40-18$

$2l+6m=22$

taking 2 common from left hand side we get

$2(l+3m)=22$

$l+3m=\frac{22}{2}$$l+3m=11$        equation$(1)$  

Now,

op=or  (radius of the circle)

$\sqrt{(l-5)^{2}+(m-(-1))^{2}  }$ = $\sqrt{(l-(-2))^{2}+(m-6)^{2}  }$

Cancel square roots from both the sides

Then we have,

$(l-5)^{2} +(m+1)^{2}$ = $(l+2)^{2} +(m-6)^{2}$

$l^{2}-10l+25+m^{2}+2m+1$ =  $l^{2}+4l+4+m^{2}-12m+36$

Now cancelling on both the sides, we get

$-10l+2m+26$ =  $4l-12m+40$

$-10l+2m-4l+12m=40-26$

$-14l+14m=14$

Taking 14 common on left hand side

$14(-l+m)=14$

$-l+m=1$       equation$(2)$          

Adding  equation $(1)$ and $(2)$ , we get

$l-l+3m+m=11+1$

$4m=12$

$m=\frac{12}{4}$

$m=3$

Substituting value of m in equation$(1)$

$l+3(3)=11$

$l=11-9$$l=2$

Hence circumcentre is o(l,m) = o(3,2).

Now with the help centre o(3,2) and the vertex p(5,-1)

radius of the circumcircle = $\sqrt{(3-5)^{2}+(2-(-1))^{2}  }$

=$\sqrt{4+9}$

=$\sqrt{13}$