# ivyrianne's question at Yahoo! Answers regarding finding a circle

#### MarkFL

Staff member
Here is the question:

Determine the equation of a circle having its center on the line 4x+3y=2?

continuation: and tangent to the line x+y+4=0 and 7x-y+4=0
I have posted a link there to this topic so the OP can see my work.

#### MarkFL

Staff member
Hello again ivyrianne,

Let's let the equation of the circle be:

$$\displaystyle (x-h)^2+(y-k)^2=r^2$$

We have 3 parameters to determine: $h,k,r$.

In order for the center of the circle to lie on the line $4x+3y=2$, we require:

(1) $$\displaystyle 4h+3k=2$$

Now, let's rewrite the tangent lines in slope-intercept form:

a) $$\displaystyle y=-x-4$$

b) $$\displaystyle y=7x+4$$

If we have a point $\left(x_0,y_0 \right)$ and a line $y=mx+b$, then the perpendicular distance $d$ from the point to the line is given by:

$$\displaystyle d=\frac{\left|mx_0+b-y_0 \right|}{\sqrt{m^2+1}}$$

You may find derivations of this formula here:

http://www.mathhelpboards.com/f49/finding-distance-between-point-line-2952/

Using this formula for $d=r$ and the two tangent lines, we find that we require:

$$\displaystyle r=\frac{|-h-4-k|}{\sqrt{2}}=\frac{|7h+4-k|}{5\sqrt{2}}$$

Using the definition $$\displaystyle |x|\equiv\sqrt{x^2}$$, we may square through and write:

$$\displaystyle \left(5(h+k+4) \right)^2=(7h-k+4)^2$$

Solving (1) for $k$, we find:

$$\displaystyle k=\frac{2-4h}{3}$$

Substituting, there results:

$$\displaystyle \left(5\left(h+\frac{2-4h}{3}+4 \right) \right)^2=\left(7h-\frac{2-4h}{3}+4 \right)^2$$

Combining like terms, we find:

$$\displaystyle \left(\frac{5}{3}(h-14) \right)^2=\left(\frac{5}{3}(5h+2) \right)^2$$

Hence, we must have:

$$\displaystyle (h-14)^2-(5h+2)^2=0$$

$$\displaystyle (6h-12)(-4h-16)=0$$

$$\displaystyle -24(h-2)(h+4)=0$$

$$\displaystyle h=-4,\,2$$

Case 1: $$\displaystyle h=-4$$

$$\displaystyle k=\frac{2-4(-4)}{3}=6$$

$$\displaystyle r=\frac{\left|-(-4)-4-6 \right|}{\sqrt{2}}=3\sqrt{2}$$

Case 2: $$\displaystyle h=2$$

$$\displaystyle k=\frac{2-4(2)}{3}=-2$$

$$\displaystyle r=\frac{\left|-2-4-(-2) \right|}{\sqrt{2}}=2\sqrt{2}$$

Thus, the two possible circles are:

$$\displaystyle \left(x+4 \right)^2+\left(y-6 \right)^2=18$$

$$\displaystyle \left(x-2 \right)^2+\left(y+2 \right)^2=8$$

Here is a plot of the three lines and the two circles: 