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
B
2
C
32
D
22

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.$$D=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}$$$$=\sqrt{2^{2}+2^{2}}$$$$=2\sqrt{2}$$.Maths

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