first find the general form of the perpandicular to the circle so you have ax+by=c.
rearrange for y= whatever
substitute y into your circle and you want the discriminant to be equal to 0, so that for the x there in only once y possible (the discriminant should be a quadractic in c so solve this equation)
for this second one find the formula of the line first and you should get y=kx.
substitute this in to find the two possible values of k using a similar method
(Original post by Sgany)
I am having a fair bit of trouble getting the correct answers for a few questions in my math book
Find the equations of the tangents to the circle x^2+y^2+6x-2y-15=0 that are perpendicular to the line 4x+3y+5=0
Is line 4x+3y+5=0 a tangent to the circle? If so then a line perpendicular to it would go through the centre of the circle. You can use the equation of the circle to find that using completing the square. Then you just need to work out the gradient of the tangent to the circle and insert it all into y-y1=m(x-x1).
Is this what you're looking for?