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Question

Common chord of the circles $$ \displaystyle x^{2}+y^{2}-6x-4y+9=0 $$ and $$ \displaystyle x^{2}+y^{2}-8x-6y+23=0 $$ is a diameter of the second circle then its length is


A
42
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B
2
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C
32
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D
22
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Solution

The correct option is D $$ \displaystyle 2\sqrt{2} $$
The equation of the common chord is 
$$S_{2}-S_{1}=0$$
$$(x^{2}+y^{2}-6x-4y+9)-(x^{2}+y^{2}-8x-6y+23)=0$$

$$2x+2y-14=0$$

$$x+y=7$$...(i)
Now this is the diameter of $$x^{2}+y^{2}-8x-6y+23=0$$
Hence it should cut the circle at two distinct points.
Therefore 
$$x^{2}+(7-x)^{2}-8x-6(7-x)+23=0$$
$$x^{2}+x^{2}-14x+49-8x-42+6x+23=0$$

$$2x^{2}+x(-14-8+6)+49-42+23=0$$

$$2x^{2}-16x+30=0$$
$$x^{2}-8x+15=0$$
$$x=5,3$$
$$y=2,4$$ repectively.
$$D=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}$$

$$=\sqrt{2^{2}+2^{2}}$$

$$=2\sqrt{2}$$.

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