Question 1183327
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The circle is tangent to the line 3x-4y-4=0 at the point (-4,-4).  The equation of that line in slope-intercept form is y=(3/4)x-1.<br>
To be tangent to y=(3/4)x-1 at (-4,-4), the center of the circle must be on the line perpendicular to y=(3/4)x-1 passing through (-4,-4).  A line perpendicular to y=(3/4)x-1 has slope (-4/3).  Find the equation of the line with slope -4/3 passing through (-4,-4).<br>
{{{y = mx+b}}}
{{{-4 = (-4/3)(-4)+b}}}
{{{-4 = 16/3+b}}}
{{{b=-4-16/3=-12/3-16/3=-28/3}}}<br>
The equation is {{{y=(-4/3)x-28/3}}}<br>
It is given that the center of the circle is also on the line x+y+7=0, or y=-x-7.<br>
We have the equations of two lines, both of which must contain the center of the circle.  So find the center of the circle by finding the intersection of those two lines.<br>
{{{-x-7=(-4/3)x-28/3}}}
{{{-3x-21=-4x-28}}}
{{{x=-7}}}
{{{y = -x-7=7-7=0}}}<br>
The center of the circle is at (-7,0).<br>
The radius of the circle is then the distance from the center (-7,0) to the given point of tangency (-4,-4).  A quick sketch, or the distance formula, will show the radius to be 5.<br>
The equation of the circle with center (h,k) and radius r is<br>
{{{(x-h)^2+(y-k)^2=r^2}}}<br>
{{{(x+7)^2+(y-0)^2=5^2}}}
{{{(x+7)^2+y^2=25}}}<br>
ANSWER: The equation of the circle is {{{(x+7)^2+y^2=25}}}<br>
A graph....
red line: 3x-4y-4=0, or y=(3/4)x-1, to which the circle is tangent
green line: (-4/3)x-28/3, perpendicular to the red line passing through (-4,-4)
blue line: x+y+7=0, or y=-x-7, on which the center of the circle must lie
the center of the circle is the intersection of the green and blue lines, (-7,0)<br>
{{{graph(400,400,-10,2,-10,2,(3/4)x-1,(-4/3)x-28/3,-x-7,sqrt(25-(x+7)^2),-sqrt(25-(x+7)^2))}}}<br>