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

3 Answers
Dec 18, 2017

Answer:

#(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

Answer:

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

Answer:

# 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.#