Question

The equation of the circle having centre (1, –2) and passing through the point of intersection of lines 3x + y = 14, 2x + 5y = 18 is

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

The point of intersection of 3x + y - 14 = 0 and 2x + 5y - 18 = 0 are
$x=\frac{-18+70}{15-2},y=\frac{-28+54}{13}\Rightarrow x=4,y=2$
i.e., point is (4, 2).
Therefore radius is $\sqrt{\left(9\right)+\left(16\right)}=5$ and equation is $x^2+y^2-2x+4y-20=0$
Trick: The only circle is $x^2+y^2-2x+4y-20=0,$ whose centre is (1, -2).