Question

If the vertices of a triangle be $( 2 , - 2 )$, $( - 1 , - 1 )$ and (5, 2), then the equation of its circumcircle is.

• A. $x ^ { 2 } + y ^ { 2 } + 3 x + 3 y + 8 = 0$
• B. $x ^ { 2 } + y ^ { 2 } - 3 x - 3 y - 8 = 0$
• C. $x ^ { 2 } + y ^ { 2 } - 3 x + 3 y + 8 = 0$
• D. None of these

Answer 1

Answer: (B)

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

Answer 2

Let us find the equation of family of circles

through $( 2 , - 2 )$ and $( - 1 , - 1 )$.

i.e. $( x - 2 ) ( x + 1 ) + ( y + 2 ) ( y + 1 ) + \lambda \left( \frac { y + 2 } { - 2 + 1 } - \frac { x - 2 } { 2 + 1 } \right) = 0$

Now for point (5, 2) to lie on it, we should have $\lambda$given by $3 \cdot 6 + 4 \cdot 3 + \lambda \left( \frac { 4 } { - 1 } - 1 \right) = 0 \Rightarrow \lambda = \frac { 30 } { 5 } = 6$

Hence equation is

$( x - 2 ) ( x + 1 ) + ( y + 2 ) ( y + 1 ) + 6 \left( \frac { y + 2 } { - 1 } - \frac { x - 2 } { 3 } \right) = 0$

or $x ^ { 2 } + y ^ { 2 } - 3 x - 3 y - 8 = 0$.

Trick: Here the circle $x ^ { 2 } + y ^ { 2 } - 3 x - 3 y - 8 = 0$is satisfied by (2, –2), (–1, –1) and (5, 2). Therefore students must check such type of problems conversely.