A circle has equation x^2+y^2+6x+10y-7=0. Find the equation of the tangent line through the point on the circle (-8,-1).
The circle can be written (x+3)2+(y+5)^2 = 41 and so has center (-3,-5).
The gradient through the center and the point (-8,-1) is;m = (-5+1)/(-3+8) = -4/5.
The tangent line is perpendicular to this so has gradient;mt= 5/4.
Now we have, y = (5/4)x + c. To find c we use the point on the line (-8,-1);-1 = (5/4)*-8 = c, c = -1 +10 = 9.
So the final answer is y = (5/4)x + 9.
RL
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