The vertices A - B - C - D of the triangles are -2 - 1- - -5 - 2- and -3 - 4- respectively - Find the coordinates of the circumcentre and also the radius of the circumcircle

Question

Toppr Admin · Accepted Answer

Let P - x - y - be the co -ordinates of the centre of the circumcircle-xA0-x2234-PA2-PB2andPB2-PC2-x2234-3x-y-12-xA0-andx-3y-10solving these two we get the centre as -13 - 4 - 9 - 5 - Having found P we can say that-xA0-PA2-r2-50-16-x2234-r-5-x221A-2-4

The vertices A , B , C , D of the triangles are (2 , 1) , (5 , 2) and (3 , 4) respectively . Find the coordinates of the circumcentre and also the radius of the circumcircle

Let P ( x , y ) be the co -ordinates of the centre of the circumcircle
$∴ P A^{2} = P B^{2} a n d P B^{2} = P C^{2}$
$∴ 3 x + y = 12 a n d x + 3 y = 10$
solving these two we get the centre as (13 / 4 , 9 / 5 ) Having found P we can say that
$P A^{2} = r^{2} = 50 / 16 ∴ r = 5 \sqrt{2} / 4$

The vertices A , B , C , D of the triangles are (2 , 1) , (5 , 2) and (3 , 4) respectively . Find the coordinates of the circumcentre and also the radius of the circumcircle

Let P ( x , y ) be the co -ordinates of the centre of the circumcircle ∴PA2=PB2andPB2=PC2∴3x+y=12 andx+3y=10solving these two we get the centre as (13 / 4 , 9 / 5 ) Having found P we can say that PA2=r2=50/16∴r=5√2/4

Let P ( x , y ) be the co -ordinates of the centre of the circumcircle
$∴ P A^{2} = P B^{2} a n d P B^{2} = P C^{2}$
$∴ 3 x + y = 12 a n d x + 3 y = 10$
solving these two we get the centre as (13 / 4 , 9 / 5 ) Having found P we can say that
$P A^{2} = r^{2} = 50 / 16 ∴ r = 5 \sqrt{2} / 4$