The equation of the circle having centre (1, –2) and passing through

1.	The equation of the circle having centre (1, –2) and passing through the point of intersection of the lines 3x + y = 14 and 2x + 5y = 18 is(A) x2 + y2 – 2x + 4y – 20 = 0 (B) x2 + y2 – 2x – 4y – 20 = 0(C) x2 + y2 + 2x – 4y – 20 = 0 (D) x2 + y2 + 2x + 4y – 20 = 0
Answer» The correct option is (A). The point of intersection of 3x + y – 14 = 0 and 2x + 5y – 18 = 0 are x = 4, y = 2, i.e., the point (4, 2) Therefore, the radius is = √(9+16)=5 and hence the equation of the circle is given by (x – 1)² + (y + 2)² = 25 or, x² + y² – 2x + 4y – 20 = 0.

The equation of the circle having centre (1, –2) and passing through the point of intersection of the lines 3x + y = 14 and 2x + 5y = 18 is(A) x2 + y2 – 2x + 4y – 20 = 0 (B) x2 + y2 – 2x – 4y – 20 = 0(C) x2 + y2 + 2x – 4y – 20 = 0 (D) x2 + y2 + 2x + 4y – 20 = 0

Answer»

The correct option is (A). The point of intersection of 3x + y – 14 = 0 and 2x + 5y – 18 = 0 are x = 4, y = 2, i.e., the point (4, 2)

Therefore, the radius is = √(9+16)=5 and hence the equation of the circle is given by

(x – 1)² + (y + 2)² = 25

or, x² + y² – 2x + 4y – 20 = 0.