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Question

The equation of the circle passing through the point $$(1,1)$$ and the points of intersection of $${x}^{2}+{y}^{2}-6x-8=0$$ and $${ x }^{ 2 }+{ y }^{ 2 }-6=0$$


A
x2+y2+3x5=0
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B
x2+y24x+2=0
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C
x2+y2+6x4=0
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D
x2+y24y2=0
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Solution

The correct option is B $${ x }^{ 2 }+{ y }^{ 2 }+3x-5=0$$
Let $${S}_{1}\equiv {x}^{2}+{y}^{2}-6x-8=0$$
and $${S}_{2}\equiv { x }^{ 2 }+{ y }^{ 2 }-6=0$$
Now the equation of the circle passing through the point $$(1,1)$$ and the point of intersection of $${S}_{1}$$ and $${S}_{2}$$
$${S}_{1}+\lambda{S}_{2}=0$$
$$\Rightarrow$$ $$({x}^{2}+{y}^{2}-6x-8)+\lambda({ x }^{ 2 }+{ y }^{ 2 }-6)=0$$
$$\Rightarrow$$ $$(1+\lambda){x}^{2}+(1+\lambda){y}^{2}-6x+(-8-6\lambda)=0........(i)$$
Since eq. $$(i)$$ passes through the point $$(1,1)$$
$$\because$$ Put $$x=1, y=1$$ in eq. $$(ii)$$ we get
$$(1+\lambda)+(1+\lambda)-6+(-8-6\lambda)=0$$
$$\Rightarrow$$ $$-4\lambda-12=0$$
$$\Rightarrow$$ $$\lambda=-3$$
On putting the value of '$$\lambda$$' in Eq. $$(i)$$ we get
$$-2{x}^{2}-2{y}^{2}-6x+10=0$$
$$\Rightarrow$$ $${x}^{2}+{y}^{2}+3x-5=0$$

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