Question

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

• 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 $\left(\right.1,-2\left.\right)$ and $\left(\right.4,-3\left.\right)$ as the extremities of a diameter is given as,
$\left(\right.x-1\left.\right)\left(\right.x-4\left.\right)+\left(\right.y+2\left.\right)\left(\right.y+3\left.\right)=0$
$x^{2}+y^{2}-5x+5y+10=0\ldots \left(\right.1\left.\right)$
The equation of the line through $\left(\right.1,-2\left.\right)$ and $\left(\right.4,-3\left.\right)$ is given as,
$x+3y+5=0\ldots \left(\right.2\left.\right)$
Any circle through the intersection of $\left(\right.1\left.\right)$ and $\left(\right.2\left.\right)$ 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,
$15\left(x^{2} + y^{2}\right)-94x+18y+55=0$