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
Answered by Rhianna L. Maths tutor

1032 Views

See similar Maths Scottish Highers tutors

Related Maths Scottish Highers answers

All answers ▸

Find the stationery points of x^3 + 3x^2 - 24x + 7 and determine whether the slope is increasing or decreasing at x=3.


Solve log_2(3x + 7) = 3 + log_2(x – 1), x > 1.


Calculate the rate of change of d(t )=2/(3t), t ≠ 0, when t=6.


Differentiate 5x^2 - 7x +9


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences