Question

# Find the equation of the circle passing through the points of intersection of the circles $$x^2+y^2-4x-6y-12=0$$ and $$x^2+y^2+6x+4y-12=0$$ and intersection the circle $$x^2+y^2-2x-4=0$$ orthogonally.

Solution

## Let, the equation of the circle through the intersection of given circle is,$${ x }^{ 2 }+{ y }^{ 2 }-4x-6y+12=0\quad \quad \longrightarrow (1)$$Here, the equation of radical axis for the circles $${ x }^{ 2 }+{ y }^{ 2 }-4x-6y+12=0$$and $${ x }^{ 2 }+{ y }^{ 2 }+6x-+4y-12=0$$ is,$$-10x-10y=0$$So, equation $$(1)$$ can be written as,$${ x }^{ 2 }+{ y }^{ 2 }-4x-6y-12+\lambda (-10x-10y)=0\\ \Rightarrow { x }^{ 2 }+{ y }^{ 2 }-x(10\lambda +4)-y(10\lambda +6)-12=0$$It cuts the circle $$v$$ orthogonally.$$\therefore2g{ g }_{ 1 }+2f{ f }_{ 1 }=c+{ c }_{ 1 }\\ \Rightarrow (10\lambda +4)(1)+(10\lambda +6)(0)=-12-4\\ \Rightarrow \lambda =-2$$Hence, the required equation of the circle is,$${ x }^{ 2 }+{ y }^{ 2 }-4x-6y-12-2(-10x-10y)=0\\ \Rightarrow { x }^{ 2 }+{ y }^{ 2 }+16x+14y-12=0.$$Maths

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