question_answer

The equation of the circle of radius 5 in the first quadrant which touches x-axis and the line \[4y=3x\]is

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 \[{{(x-15)}^{2}}+{{(y-5)}^{2}}={{5}^{2}}\] \[\Rightarrow \] \[{{x}^{2}}-30x+225+{{y}^{2}}-10y+225=0\]