12. Find the equation of the circle...
Question  # 12. Find the equation of the circle with radius 5 whose centre lies on r-axis and passes through the point (2,3).

JEE/Engineering Exams
Maths
Solution 105 4.0 (1 ratings)  Let the equation of the required circle be ( (x-h)^{2}+(y-k)^{2}=r^{2} ) since the radius of the circle is 5 and its centre lies on the ( x ) -axis, ( k=0 ) and ( r=5 ) Now, the equation of the circle becomes ( (x-h)^{2}+y^{2}=25 ) It is given that the circle passes through point (2,3) . ( therefore(2-h)^{2}+3^{2}=25 ) ( Rightarrow(2-h)^{2}=25-9 ) ( Rightarrow(2-h)^{2}=16 ) ( Rightarrow 2-h=pm sqrt{16}=pm 4 ) If ( 2-h=4, ) then ( h=-2 ) If ( 2-h=-4, ) then ( h=6 ) When ( h=-2 ), the equation of the circle becomes ( (x+2)^{2}+y^{2}=25 ) ( x^{2}+4 x+4+y^{2}=25 ) ( x^{2}+y^{2}+4 x-21=0 ) When ( h=6, ) the equation of the circle becomes ( (x-6)^{2}+y^{2}=25 ) ( x^{2}-12 x+36+y^{2}=25 ) ( x^{2}+y^{2}-12 x+11=0 ) Quick and Stepwise Solutions Just click and Send OVER 20 LAKH QUESTIONS ANSWERED Download App for Free