A) \[{{x}^{2}}+{{y}^{2}}-24x-y-25=0\]
B) \[{{x}^{2}}+{{y}^{2}}-30x-10y+225=0\]
C) \[{{x}^{2}}+{{y}^{2}}-16x-18y+64=0\]
D) \[{{x}^{2}}+{{y}^{2}}-20x-12y+144=0\]
Correct Answer: B
Solution :
Since, the perpendicular from centre to the circle is equal to the radius of the circle. \[\therefore \] \[\frac{3(g)-4(5)}{\sqrt{{{3}^{2}}+{{4}^{2}}}}=5\] \[\Rightarrow \] \[3g=25+20\Rightarrow g=15\] \[\therefore \] Equation of circle whose centre \[(15,5)\] and radius 5 is \[{{(x-15)}^{2}}+{{(y-5)}^{2}}={{5}^{2}}\] \[\Rightarrow \] \[{{x}^{2}}-30x+225+{{y}^{2}}-10y+225=0\]You need to login to perform this action.
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