Question

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

• A. $x ^ { 2 } + y ^ { 2 } - 2 x + 4 y - 20 = 0$
• B. $x ^ { 2 } + y ^ { 2 } - 2 x - 4 y - 20 = 0$
• C. $x ^ { 2 } + y ^ { 2 } + 2 x - 4 y - 20 = 0$
• D. $x ^ { 2 } + y ^ { 2 } + 2 x + 4 y - 20 = 0$

Answer 1

Answer: (A)

$x ^ { 2 } + y ^ { 2 } - 2 x + 4 y - 20 = 0$

Answer 2

The point of intersection of $3 x + y - 14 = 0$ and

$2 x + 5 y - 18 = 0$ 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 $\sqrt { ( 9 ) + ( 16 ) } = 5$ and equation is

$x ^ { 2 } + y ^ { 2 } - 2 x + 4 y - 20 = 0$ .

Trick : The only circle is $x ^ { 2 } + y ^ { 2 } - 2 x + 4 y - 20 = 0$ , whose centre is (1, –2­).