Equation of the Circle (Part 2)

[mathdad]

Find the equation of the circle tangent to the y-axis and with center (3, 5).

Can someone provide the steps needed to solve this problem?

 

[MarkFL]

The equation of a circle centered ar $(h,k)$ is given by:

$$(x-h)^2+(y-k)^2=r^2$$

If the circle is tangent to the $y$-axis, then its radius must be $r=|h|\implies r^2=h^2$, thus we have:

$$(x-h)^2+(y-k)^2=h^2$$

We are given $(h,k)=(3,5)$, so plug in those numbers. :D

 

[mathdad]

The equation of a circle centered ar $(h,k)$ is given by:

$$(x-h)^2+(y-k)^2=r^2$$

If the circle is tangent to the $y$-axis, then its radius must be $r=|h|\implies r^2=h^2$, thus we have:

$$(x-h)^2+(y-k)^2=h^2$$

We are given $(h,k)=(3,5)$, so plug in those numbers. :D

(x - h)^2 + (y - k)^2 = h^2

(x - 3)^2 + (y - 5)^2 = 3^2

(x - 3)^2 + (y - 5)^2 = 9