document.write( "Question 608398: Find the equation of the tangent to the curve (x-3)^2+(y+2)^2=25 at the point on the curve, in the fourth quadrant, where x=7. I've already solved for the formula, but how do you find the y value for where x=7? Do you have to use the distance formula from the C(3,-2) to P(7, y) and solve for y or what? I'm confused. D: \n" ); document.write( "
Algebra.Com's Answer #383201 by lwsshak3(11628)\"\" \"About 
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Find the equation of the tangent to the curve (x-3)^2+(y+2)^2=25 at the point on the curve, in the fourth quadrant, where x=7.
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\n" ); document.write( "Find y using equation of curve (circle)
\n" ); document.write( "Plug in 7 for x
\n" ); document.write( "(7-3)^2+(y+2)^2=25
\n" ); document.write( "(4)^2+(y+2)^2=25
\n" ); document.write( "16+(y+2)^2=25
\n" ); document.write( "(y+2)^2=25-16=9
\n" ); document.write( "take sort of both sides
\n" ); document.write( "y+2=±√9=±3
\n" ); document.write( "y=-2±3
\n" ); document.write( "y=-5
\n" ); document.write( "or
\n" ); document.write( "y=1 (not in 4th quadrant, so y=-5)
\n" ); document.write( "Point of tangency: (7,-5)
\n" ); document.write( "..
\n" ); document.write( "Finding slope of line tangent to circle at (7,-5)
\n" ); document.write( "This line is perpendicular to the line connecting center of circle(3,-2) to point of tangency (7,-5)
\n" ); document.write( "So its slope is the negative reciprocal of the tangent line slope.
\n" ); document.write( "slope of perpendicular line=change in y/change in x=∆y/∆x=(-2-(-5))/(3-7)=-3/4
\n" ); document.write( "slope of tangent line=4/3
\n" ); document.write( "Equation of line: y=mx+b, m=slope, b=y-intercept
\n" ); document.write( "y=4x/3+b
\n" ); document.write( "solve for b using coordinates of point of tangency(7,-5)
\n" ); document.write( "-5=4*7/3+b
\n" ); document.write( "b=-43/3
\n" ); document.write( "equation of line tangent to circle where x=7: y=4x/3-43/3
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