Question

What is the equation of a circle with center (1,3) and containing the point (1,0)?

Answer 1

eqn of circle is,
`x^2-2x+y^2-6y+1=0`

Explanation:

centre of the circle lies on pt. O(1,3)
pt. on the circumference of the circle is A(1,0)
eqn of circle is,
`(x-h)^2+(y-k)^2=r^2`
where, (x,y) is the pt. on the circle=A(1,0)
&, (h,k) is the centre of the circle=O(1,3)
now, calculating the radius of the circle
which is,
`(1-1)^2+(0-3)^2=r^2`
`r^2=9`
`r^cancel2=sqrt9=3`
so, now the eqn of circle,
`(x-1)^2+(y-3)^2=(3)^2`
`x^2+1-2x+y^2+9-6y=9`
`x^2-2x+y^2-6y+1=0`