Question

Find the equation of the circle with 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 $AB=m=\frac{y_2-y_1}{x_2-x_1}=\frac{-2-1}{-1-1}=\frac{3}{2}$

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

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

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

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

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

$2y+1=3x$

$3x-2y=1.........\left(1\right)$

And given equation of line is

$x+2y=1$ $\dots ....\left(2\right)$

By equation (1) and (2) to, we get

$3x-2y=1$

$x+2y=1$

$4x=2$

$x=\frac{1}{2}$

Put the value of $x$ in equation $\left(2\right)$ and we get,

Then $y=\frac{1}{4}$

Now, the centre of circle is at $(\frac{1}{2},\frac{1}{4})$.

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

Then, radius of circle is $=\sqrt{{(-1-\frac{1}{2})}^2+{(-2-\frac{1}{4})}^2}=\frac{\sqrt{117}}{4}$

Therefore, the equation of circle is

${(x-\frac{1}{2})}^2+{(y-\frac{1}{4})}^2=\frac{117}{16}$