Question

# A circle passes through the point of intersction of circles $${ x }^{ 2 }+{ y }^{ 2 }-6x+2y+4=0$$ and $${ x }^{ 2 }+{ y }^{ 2 }+2x-4y-6=0$$ and its centre lies on the line $$y=x$$. Its equation will be

A
7(x2+y2)10x10y12=0
B
7(x2+y2)10x10y1=0
C
x2+y210x10y12=0
D
none of these

Solution

## The correct option is B $$7({ x }^{ 2 }+{ y }^{ 2 })-10x-10y-12=0$$A circle passes through the point of intersction of circles $${ x }^{ 2 }+{ y }^{ 2 }-6x+2y+4=0$$ and $${ x }^{ 2 }+{ y }^{ 2 }+2x-4y-6=0$$ is$${ x }^{ 2 }+{ y }^{ 2 }-6x+2y+4+k\left( { x }^{ 2 }+{ y }^{ 2 }+2x-4y-6 \right) =0$$ ....(i)$$\Rightarrow \left( 1+k \right) \left( { x }^{ 2 }+{ y }^{ 2 } \right) -2\left( 3-k \right) x+2\left( 1-2k \right) +4-6k=0$$After dividing by $$(1+k)$$ we get centre as $$\left( \dfrac { 3-k }{ 1+k } ,\dfrac { 2k-1 }{ 1+k } \right)$$ and  which lies on the line $$y=x$$Therefore, $$\dfrac { 3-k }{ 1+k } =\dfrac { 2k-1 }{ 1+k }$$$$\Rightarrow k=\dfrac 43$$, Putting this in (i) and simplifying, we getequation is circle as $$7({ x }^{ 2 }+{ y }^{ 2 })-10x-10y-12=0$$Ans: AMaths

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