Question

How to determine the general form of the equation of the circle whose center is (3,2) and whose graph contains the point (1,2)?

Answer 1

`(x-3)^2+(y-2)^2=4`

Explanation:

The general equation of a circle is:

`(x-a)^2+(y-b)^2=r^2`

where:

• `(a,b)` are the coordinates of the circle's center

• `r` is the radius of the circle

Since `(1,2)` lies on the circle, then the circle has a radius of:

`r=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

`=sqrt(2^2-0^2)`

`=sqrt4`

`=2`

So, the equation of this circle will be:

`(x-3)^2+(y-2)^2=2^2`

`(x-3)^2+(y-2)^2=4`

graph{(x-3)^2+(y-2)^2=4 [-8.89, 8.89, -4.444, 4.445]}

Answer 2

`x^2+y^2-6x-4y+9=0`

Explanation:

`"the equation of a circle in standard form is"`

`color(red)(bar(ul(|color(white)(2/2)color(black)((x-a)^2+(y-b)^2=r^2)color(white)(2/2)|)))`

`"where "(a,b)" are the coordinates of the centre and r"`
`"is the radius"`

`"here "(a,b)=(3,2)`

`"the radius is the distance from the centre to the point"`
`(1,2)" on the circle"`

`r=sqrt((3-1)^2+(2-2)^2)=sqrt4=2`

`(x-3)^2+(y-2)^2=4larrcolor(red)" in standard form"`

`"to obtain the equation in "color(blue)"general form"`

`•color(white)(x)x^2+y^2+2gx+2fy+c=0`

`"expand the standard form and rearrange into this form"`

`x^2-6x+9+y^2-4y+4-4=0`

`x^2+y^2-6x-4y+9=0larrcolor(red)"in general form"`