Question

Equation of a circle which passes through (3,6) and touches the axes is
A. `x^(2)+y^(2)+6x+6y+3=0`
B. `x^(2)+y^(2)-6x-6y-9=0`
C. `x^(2)+y^(2)-6x-6y+9=0`
D. none of these

Answer 1

Correct Answer - C
Let centre of the circle be (a,a), then equation of the circle is `(x-a)^(2)+(y-a)^(2)=a^(2)`

Since, the point (3,6) lies on the circle then
`(3-a)^(2)+(6-a)^(2)=a^(2)`
`rArra^(2)+9-6a+36-12a+a^(2)=a^(2)`
`rArra^(2)-15a-3a+45=0`
`rArr a^(2)-15a-3a+45=0`
`rArra(a-15)-3(a-15)=0`
`rArr(a-3)(a-15)=0`
`rArr a=3,a=15`
So, the equation of the circle is
`(x-3)^(2)+(y-3)^(2)=9`
`rArrx^(2)-6x+9+y^(2)-6y+9=9`
`rArr x^(2)+y^(2)-6x-6y+9=0`