Question

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

Answer 1

The general equation of the circle is-

$x^2+y^2+2gx+2fy+c=0-eq\left(1\right)$

where (-g,-f) is the center of the circle

$1+4+2g\left(1\right)+2f(-2)+c=0$

$\Rightarrow 2g-4f+c=-5-eq\left(2\right)$

$16+9+2g\left(4\right)+2f(-3)+c=0$

$\Rightarrow 25+8g-6f+c=0$

$8g-6f+c=-25-eq\left(3\right)$

Now the center of the circle will satisfy the given straight line equation because it lies on it. Hence,

$3(-g)+4(-f)=7$

$3g+4f=-7-eq\left(4\right)$

$\text{Subtracting eq(1) from eq(2)}$

$6g-2f=-20$

$\Rightarrow 3g=-10+f-eq\left(5\right)$

$\text{Putting eq(5) in eq(4)}$

$-10+f+4f=-7$

$\Rightarrow 5f=3$

$\Rightarrow f=\frac{3}{5}$

$\text{Putting the value of f in eq(4), we get}$

$3g=-10+\frac{3}{5}$

$\Rightarrow 3g=\frac{-50+3}{5}$

$\Rightarrow g=\frac{-53}{15}$

$\text{puttinng the value of f and g in eq(1)}$

$2\left(\frac{-53}{15}\right)-4\left(\frac{3}{5}\right)+c=-5$

$\Rightarrow \frac{-106}{15}-\frac{12}{5}+c=-5$

$\Rightarrow \frac{-106-36}{15}+c=-5$

$\Rightarrow c=-5+\frac{142}{15}$

$\Rightarrow c=\frac{68}{15}$

$\text{putting the values of g,f,c in the general equation of the circle, we get the equation of the required circle}$

$x^2+y^2+2\left(\frac{-53}{15}\right)x+2\left(\frac{3}{5}\right)y+\frac{68}{15}=0$

$\Rightarrow 15x^2+15y^2-106x+30y+68=0$