Question

Find the equation of the circle passing through the points $(1,1),(-1,-2)$ and whose centre lies on the line $x+2y=1$

Answer 1

Since the circle passes through given points $A(x_1,y_1)=(1,1)$ and $B(x_2,y_2)=(-1,-2)$

i.e. that the centre of circle is equidistant from the points A and B.

The slope of line $AB=C=m=\frac{y_2-y_1}{x_2-x_1}=\frac{-2-1}{-1-1}=\frac{3}{2}$

Mid point of line $AB=$ $(x^{{}^{\prime }},y^{{}^{\prime }})=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

$(x^{{}^{\prime }},y^{{}^{\prime }})=(\frac{1-1}{2},\frac{1-2}{2})$

$(x^{{}^{\prime }},y^{{}^{\prime }})=(0,\frac{-1}{2})$

Slope of perpendicular bisector of $AB$ ie $C=m^{\prime }=\frac{-1}{m}=\frac{-2}{3}$

So, equation of line $CO$ is $y-y^{{}^{\prime }}=m^{\prime }(x-x^{{}^{\prime }})$

$y-\frac{-1}{2}=\frac{-2}{3}(x-0)$

$6y+3=-4x$

$4x+6y=-3.........\left(1\right)$

And given equation of line is

$x+2y=1$ ….... (2)

By subtracting equations (2)*3 from (1), we get

$4x+6y=-3$

$-(3x+6y=3)$

$x=-6$

put the value of $x$ in equation (2) and we get,

then $y=\frac{7}{2}$

now, the centre of circle is at $(-6,\frac{7}{2})$.

Since the circle passes through $(-1,-2)$.

Then, radius of circle is $=\sqrt{{(-6+1)}^2+{(\frac{7}{2}+2)}^2}=\frac{\sqrt{221}}{2}$

Hence the equation of circle is

${(x+6)}^2+{(y-\frac{7}{2})}^2={\left(\frac{\sqrt{221}}{2}\right)}^2$

$x^2+12x+36+y^2-7y+\frac{49}{4}=\frac{221}{4}$

$x^2+12x+36+y^2-7y+\frac{-172}{4}=0$

$x^2+12x+36+y^2-7y-43=0$