Question 1068273
Find the line perpendicular to the tangent line that goes through the circle center. 
{{{3x+4y+34=0}}}
{{{4y=-3x-34}}}
{{{y=-(3/4)x-17/2}}}
Perpendicular lines have slopes that are negative reciprocals.
{{{-(3/4)m[2]=-1}}}
{{{m[2]=4/3}}}
Let the circle center be located at (h,k) and let the intersection point of the tangent line and the perpendicular line be (a,b).
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*[illustration FCX4.png].
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So you know that,
1.{{{3a+4b+34=0}}}
2.{{{4h+3k+7=0}}}
The perpendicular line goes through both (h,k) and (a,b) so,
3.{{{k-b=(4/3)(h-a)}}}
and from the distance formula,
{{{4^2=(h-a)^2+(k-b)^2}}}
4.{{{(h-a)^2+(k-b)^2=16}}}
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Substitute 3 into 4,
{{{(h-a)^2+(16/9)(h-a)^2=16}}}
{{{(25/9)(h-a)^2=16}}}
{{{(h-a)^2=(16*9)/25}}}
{{{h-a= 0 +- 12/5}}}
First case,
{{{h=a + 12/5}}}
So then going back through the equations and substituting for h,
5. {{{3a+4b=-34}}}
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{{{4h+3k+7=0}}}
{{{4(a+12/5)+3k=-7}}}
{{{4a+3k=-7-12/5}}}
6.{{{4a+3k=-83/5}}}
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{{{k-b=(4/3)(12/5)}}}
7.{{{k-b=16/5}}}
You have three equations, three unknowns, using Cramer's rule,
{{{a=-2/5}}}
{{{h=-41/5}}}
{{{k=-5}}}
and then,
{{{h=-2/5 + 12/5}}}
{{{h=10/5}}}
{{{h=2}}}
So the first circle is,
{{{highlight((x-2)+(y+5)^2=16)}}}
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Second case,
{{{h=a - 12/5}}}
Again substituting as needed for h,
8. {{{3a+4b=-34}}}
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{{{4(a-12/5)+3k+7=0}}}
{{{4a+3k=-7+48/5}}}
9.{{{4a+3k=13/5}}}
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{{{k-b=(4/3)(-12/5)}}}
10.{{{k-b=-16/5}}}
Again you have three equations, three unknowns, using Cramer's rule,
{{{a=754/35}}}
{{{h=-863/35}}}
{{{k=-195/7}}}
and then,
{{{h=a - 12/5}}}
{{{h=754/35-12/5}}}
{{{h=134/7}}}
So the second circle is,
{{{highlight((x-134/7)+(y+195/7)^2=16)}}}
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*[illustration gd11.JPG].