A) \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-4x-6y+8z-20=0\]
B) \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+4x-6y-8z-20=0\]
C) \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-4x-6y-8z+20=0\]
D) None of the above
Correct Answer: A
Solution :
Given, centre of sphere is\[(2,3,-4)\]which touches the plane\[2x+6y-3z+15=0\]. So, perpendicular distance from the centre to the plane, \[p=\frac{2\times 2+6\times 3-3\times (-4)+15}{\sqrt{{{(2)}^{2}}+{{(6)}^{2}}+{{(-3)}^{2}}}}\] \[=\frac{22+12+15}{\sqrt{4+36+9}}=\frac{49}{\sqrt{49}}=\frac{49}{7}\] \[\Rightarrow \] \[p=7\] Hence, equation of the sphere is \[{{(x-2)}^{2}}+{{(y-3)}^{2}}+{{(z+4)}^{2}}={{(7)}^{2}}\] \[\Rightarrow \]\[{{x}^{2}}+4-4x+{{y}^{2}}+9-6y+{{z}^{2}}+16+8z=49\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-4x-6y+8z+29-49=0\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-4x-6y+8z-20=0\]
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