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I have posted a link there to this topic so the OP can see my work.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

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I have posted a link there to this topic so the OP can see my work.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

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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: