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

3 Answers
Dec 18, 2017

(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:
enter image source here

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 (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:

enter image source here

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:

enter image source here

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

Dec 18, 2017

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.