Question

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

Answer 1

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

Explanation:

The equation for a circle is `(x-h)^2+(y-k)^2=r^2`, where `(h,k)` is the center and `r` is the radius.
Plugging in `3` and `-7` for `h` and `k`, we get `(x-3)^2+(y+7)^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: `(x-3)^2+(y+7)^2=9`

Here's the graph:

Answer 2

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 `(x - 3)^2 + (y + 7)^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 `(x - 3)^2 + (y + 7)^2 = r^2`, because the center of our circle is at `(3, -7)`.

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 `(0, -7)` on the circle, so substitute for `x = 0` and `y = -7`:

`(x - 3)^2 + (y + 7)^2 = r^2 rarr (0 - 3)^2 + (-7 + 7)^2 = r^2`

And simplify:

`(-3)^2 + (0)^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:

`(x - 3)^2 + (y + 7)^2 = r^2 rarr (x - 3)^2 + (y + 7)^2 = 3^2`

`(x - 3)^2 + (y + 7)^2 = 9`

And here's what it looks like:

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

Answer 3

`x^2+y^2-6x+14y+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 `(3,-7)` to the `Y-` Axis, is `|3|`.

`:. r=3.`

Hence, the eqn. follows : `(x-3)^2+(y+7)^2=3^2, i.e.,`

`x^2+y^2-6x+14y+49=0.`