# How do you write an equation for a circle given center in the second quadrant, tangent to y=-1, y=9 and the y-axis?

##### 2 Answers

#### Answer:

#### Explanation:

If the circle is tangent to

then it has a diameter of

which implies a radius of

If the bottom of the circle is at

the y-coordinate of the center of the circle must be

If the circle has a radius of

then the x-coordinate of the center of the circle is

So the center of the circle has coordinates

and its radius is

The standard equation for a circle is

for a circle with center

Therefore the equation of the desired circle is

graph{(x+5)^2+(y-4)^2=25 [-13.25, 9.25, -1.705, 9.545]}

#### Answer:

#### Explanation:

We know that, the **General Eqn.** of a **Circle** is, given by,

Since, **Axis,** i.e.,

Similarly,

Then, by

Also,

But,

lies in the **First Quadrant,** contrary to the **Hypo.**

as **Respected Alan P. Sir** has already **derived!**