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. $x^2+y^2-2x+4y-20 = 0$
• B. $x^2+y^2-2x+4y = 0$
• C. $x^2+y^2-2x+4y-10 = 0$
• D. $x^2+y^2+2x+4y-20 = 0$

Answer 1

Answer: (A)

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

Answer 2

First, solve the system of linear equations $3x + y = 14$ and $2x + 5y = 18$ to find the point of intersection. This point is (4, 2). The center of the circle is given as (1, -2). The radius squared ($r^2$) is the distance squared between the center (1, -2) and the point of intersection (4, 2): $r^2 = (4-1)^2 + (2-(-2))^2 = 3^2 + 4^2 = 9 + 16 = 25$. The equation of a circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$. Substituting the center (1, -2) and $r^2 = 25$, we get $(x-1)^2 + (y+2)^2 = 25$, which simplifies to $x^2 + y^2 - 2x + 4y - 20 = 0$.