Question

please help find the equations of the tangents to these circles at the points given x^2 + y^2 +2x + 4y - 12 = 0 (3, -1)

Answer 1

the circle will have a tangent at x = 3 which is in the top half of the circle

as the centre is (-1, -2)

you need to get the equation of the circle with y as the subject

(x + 1)^2 + (y + 2)^2 = 12 - 1 - 4

or

(x +1)^2 + (y + 2)^2 = 7

(y + 2)^2 = 7 - (x + 1)^2

\[y + 2 = \pm \sqrt{7 - (x + 1)^2}\]

so

\[y = \pm \sqrt{7 - (x + 1)^2} - 2\]

so you need to differentiate the which is the upper semicircle

\[y = \sqrt{7 - (x + 1)^2} - 2\]


and then substitute the find the slope of the tangent...

once you have the slope...

then use the point (3, -1) to find the equations...

hope it helps.

Answer 2

i cam't differentiate as we haven't learnt it yet ?