If the vertices of a triangle are (2, -2), (-1, -1) and (5, 2) th… - Vidyadip

MCQ

If the vertices of a triangle are (2, -2), (-1, -1) and (5, 2) then the equation of its circumcircle is:

A
$x^2+y^2+3 x+3 y+8=0$
✓
$x^2+y^2-3 x-3 y-8=0$
C
$x^2+y^2-3 x+3 y+8=0$
D
None of these

✓

Answer

Correct option: B.

$x^2+y^2-3 x-3 y-8=0$

$x^2+y^2-3 x-3 y-8=0$

Solution:
To find circumcentre we write the equation of perpendicular bisectors of two sides and find their intersection,
3x - y - 3 = 0 and 6x + 8y - 21 = 0
Their intersection point is $\big(\frac{3}{2},\frac{3}{2}\big)$
Radius of circumcircle = Distance of $\big(\frac{3}{2},\frac{3}{2}\big)$
from (2,-2) or any other vertex $=\frac{5}{\sqrt2}$
So equation of circle $=(\text{x}-\frac{3}{2})^2+(\text{y}-\frac{3}{2})^2 = \frac{25}{2}$

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