Question

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

Solution

## Since the circle passes through given  points $$A\left( {{x}_{1}},{{y}_{1}} \right)=\left( 1,1 \right)$$  and $$B\left( {{x}_{2}},{{y}_{2}} \right)=\left( -1,-2 \right)$$ i.e. that the centre of circle is equidistant from the points A and B. The slope of line  $$AB=C=m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\dfrac{-2-1}{-1-1}=\dfrac{3}{2}$$ Mid point of line $$AB=$$ $$\left( {{x}^{'}},{{y}^{'}} \right)=\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$$ $$\left( {{x}^{'}},{{y}^{'}} \right)=\left( \dfrac{1-1}{2},\dfrac{1-2}{2} \right)$$ $$\left( {{x}^{'}},{{y}^{'}} \right)=\left( 0,\dfrac{-1}{2} \right)$$Slope of perpendicular bisector of $$AB$$ ie $$C = m' = \dfrac{-1}{m}= \dfrac{-2}{3}$$ So, equation of line $$CO$$ is $$y-{{y}^{'}}=m'\left( x-{{x}^{'}} \right)$$ $$y-\dfrac{-1}{2}=\dfrac{-2}{3}\left( x-0 \right)$$ $$6y+3=-4x$$ $$4x+6y=-3\,\,\,\,\,\,.........\,\,\,\,\,\,\left( 1 \right)$$ 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   $$y=\dfrac{7}{2}$$ now, the centre of circle is at $$\left(-6,\dfrac{7}{2} \right)$$. Since the circle passes through $$\left( -1,-2 \right)$$. Then, radius of circle is $$=\sqrt{{{\left( -6+1 \right)}^{2}}+{{\left( \dfrac{7}{2}+2 \right)}^{2}}}=\dfrac{\sqrt{221}}{2}$$ Hence the equation of circle is $${{\left( x+6 \right)}^{2}}+{{\left( y-\dfrac{7}{2} \right)}^{2}}={\left({\dfrac{\sqrt{221}}{2}}\right)^{2}}$$$$x^2+12x+36 + y^2 - 7y + \dfrac{49}{4} = \dfrac{221}{4}$$$$x^2+12x+36 + y^2 - 7y + \dfrac{-172}{4} = 0$$ $$x^2+12x+36 + y^2 - 7y -43 = 0$$Mathematics

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