# What is the equation of the circle which touches the y-axis at ( 0 , 3 ) and passes through ( 1 , 0 )?

Jan 7, 2016

${\left(x - 5\right)}^{2} + {\left(y - 3\right)}^{2} = 25$

#### Explanation:

If the circle is tangent to the Y-axis at $\left(0 , 3\right)$ then its center is along the line $y = 3$.
If it passes through the point $\left(1 , 0\right)$ then its center is in Quadrant I at some point $\left({x}_{c} , 3\right)$ and its radius is ${x}_{c}$ By the Pythagorean Theorem, the square of its radius is
$\textcolor{w h i t e}{\text{XXX}} {r}^{2} = {x}_{c}^{2} = {3}^{3} + {\left({x}_{c} - 1\right)}^{2}$

Simplifying we have
$\textcolor{w h i t e}{\text{XXX}} \cancel{{x}_{c}^{2}} = 9 + \cancel{{x}_{c}^{2}} - 2 {x}_{c} + 1$
or
$\textcolor{w h i t e}{\text{XXX}} r = {x}_{c} = 5$

Using the general standard formula for a circle with radius $r$ and center $\left(a , b\right)$:
$\textcolor{w h i t e}{\text{XXX}} {\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$
and substituting $\left(a , b\right) = \left(5 , 3\right)$ for the center and $r = 5$ for the radius:
$\textcolor{w h i t e}{\text{XXX}} {\left(x - 5\right)}^{2} + {\left(y - 3\right)}^{2} = 25$