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

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

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Answer 1 · 2017-12-18T16:21:13

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

Explanation:

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

Since you want the circle to be tangent to the $y$ $y$ -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 $y$ $y$ -axis.

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

Here's the graph:
enter image source here

Answer 2 · 2017-12-18T16:26:58

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

Explanation:

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

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

If it had a radius of $1$ $1$ (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 $y$ $y$ -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 $y$ $y$ -axis (has an $x$ $x$ value of $0$ $0$ ).

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

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

${(x - 3)}^{2} + {(y + 7)}^{2} = r^{2} \to {(0 - 3)}^{2} + {(- 7 + 7)}^{2} = r^{2}$ $(x - 3)^2 + (y + 7)^2 = r^2 rarr (0 - 3)^2 + (-7 + 7)^2 = r^2$

And simplify:

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

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

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

$\pm \sqrt{9} = r$ $pm sqrt(9) = r$

$r = \pm 3$ $r = pm 3$

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

$r = 3$ $r = 3$

Plugging that back in to our equation:

${(x - 3)}^{2} + {(y + 7)}^{2} = r^{2} \to {(x - 3)}^{2} + {(y + 7)}^{2} = 3^{2}$ $(x - 3)^2 + (y + 7)^2 = r^2 rarr (x - 3)^2 + (y + 7)^2 = 3^2$

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

And here's what it looks like:

enter image source here

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

Answer 3 · 2017-12-18T16:52:02

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

Explanation:

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

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

Centre to the tangent line equals radius $r .$ $r.$

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

$:. r=3.$

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

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

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

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