Question

Find the equation of the circle whose center lies on $3x+4y=7$ and which passes through the points $(1,-2)$ and $(4,-3)$ .

• A. $15\left(x^2+y^2\right)-90x+18y+50=0$
• B. $x^2+y^2-94x+18y+55=0$
• C. $15\left(x^2+y^2\right)-94x+18y+55=0$
• D. $15\left(x^2+y^2\right)+94x+18y+55=0$

Answer 1

Answer: (C)

The equation of the circle with $(1,-2)$ and $(4,-3)$ as the extremities of a diameter is given as,
$(x-1)(x-4)+(y+2)(y+3)=0$
$x^2+y^2-5x+5y+10=0\dots (1)$
The equation of the line through $(1,-2)$ and $(4,-3)$ is given as,
$x+3y+5=0\dots (2)$
Any circle through the intersection of $(1)$ and $(2)$ can be given as,
$S+kL=0$
Since this center lies on $3x+4y=7$ , we get,
$k=-\frac{19}{15}$
$\therefore$ The required equation of the circle is given as,
$15\left(x^2+y^2\right)-94x+18y+55=0$