# How do you write the equation of a circle given center (3,-7) and tangent to the y-axis?

Dec 18, 2017

${\left(x - 3\right)}^{2} + {\left(y + 7\right)}^{2} = 9$ is a circle with its center at $\left(3 , - 7\right)$ and a radius of $3$, so the circle touches the $y$-axis at one point.

#### Explanation:

The equation for a circle is ${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$, where $\left(h , k\right)$ is the center and $r$ is the radius.
Plugging in $3$ and $- 7$ for $h$ and $k$, we get ${\left(x - 3\right)}^{2} + {\left(y + 7\right)}^{2} = {r}^{2}$

Since you want the circle to be tangent to the $y$-axis, and we know that tangent refers to a line that touches something at exactly one point, we want the circle to have a radius that will make it just big enough to reach the $y$-axis.

A radius of 3, since the $x$-value of the center is at $h = 3$, will just make the circle the right size to touch the $y$-axis. Plug $3$ in for $r$, and you get the final equation: ${\left(x - 3\right)}^{2} + {\left(y + 7\right)}^{2} = 9$

Here's the graph:

Dec 18, 2017

Using the general equation of a circle, set up the translations accordingly, then with a point on the $y$-axis with the same $y$ value as the center, solve for $r$, to finally get ${\left(x - 3\right)}^{2} + {\left(y + 7\right)}^{2} = 9$.

#### Explanation:

The equation of a circle in the center is ${x}^{2} + {y}^{2} = {r}^{2}$, where $r$ is the radius, by the Pythagorean theorem.

The center can be translated by subtracting from the $x$ and $y$ values accordingly, so in this case our equation would be ${\left(x - 3\right)}^{2} + {\left(y + 7\right)}^{2} = {r}^{2}$, because the center of our circle is at $\left(3 , - 7\right)$.

If it had a radius of $1$ (although we don't know what it is yet), this is what it looks like:

Now, as for the radius, and the circle being tangent to the y-axis. This means it has to touch the $y$-axis at some point. In other words, the size of the radius should be set in a way such that there is exactly one point on the circle that is on the $y$-axis (has an $x$ value of $0$).

Thinking about the center of the circle, and perhaps the figure above, as we expand the radius, the first point that touches the $y$-axis should have the same $y$-value as the center of the circle, which in this case, has a $y$-value of $- 7$.

Why don't we plug that into the equation we already have, and solve for $r$? We need a radius $r$ such that there exists a point $\left(0 , - 7\right)$ on the circle, so substitute for $x = 0$ and $y = - 7$:

${\left(x - 3\right)}^{2} + {\left(y + 7\right)}^{2} = {r}^{2} \rightarrow {\left(0 - 3\right)}^{2} + {\left(- 7 + 7\right)}^{2} = {r}^{2}$

And simplify:

${\left(- 3\right)}^{2} + {\left(0\right)}^{2} = {r}^{2}$

$9 + 0 = {r}^{2}$

$9 = {r}^{2}$

$\pm \sqrt{9} = r$

$r = \pm 3$

That makes sense algebraically, but since this is a geometric figure, radii are positive:

$r = 3$

Plugging that back in to our equation:

${\left(x - 3\right)}^{2} + {\left(y + 7\right)}^{2} = {r}^{2} \rightarrow {\left(x - 3\right)}^{2} + {\left(y + 7\right)}^{2} = {3}^{2}$

${\left(x - 3\right)}^{2} + {\left(y + 7\right)}^{2} = 9$

And here's what it looks like:

Indeed, the center is at $\left(3 , - 7\right)$ and it touches the $y$-axis at exactly one point: $\left(0 , - 7\right)$.

Dec 18, 2017

${x}^{2} + {y}^{2} - 6 x + 14 y + 49 = 0.$

#### Explanation:

The reqd. circle touches the $Y -$ Axis.

From Geometry, we know that, the $\bot -$ distance from

Centre to the tangent line equals radius $r .$

Now, the $\bot -$ distance from $\left(3 , - 7\right)$ to the $Y -$ Axis, is $| 3 |$.

$\therefore r = 3.$

Hence, the eqn. follows : ${\left(x - 3\right)}^{2} + {\left(y + 7\right)}^{2} = {3}^{2} , i . e . ,$

${x}^{2} + {y}^{2} - 6 x + 14 y + 49 = 0.$