Point K(8,-5) lies on the circle x^2 +y^2 - 12x - 6y - 23. find the equation of the tangent at K.

First, find the centre of the circle. Since we know the equation of the circle, we know that the centre (-g,-f) comes from the part of the equation -12x-6y, where -12=2g and -6=2f. Therefore, to get the centre coordinate, we divide both numbers by -2 to get centre (6,3).
The, we can calculate the gradient of the radius between the centre and point K using the formula (y₂-y₁)/(x₂-x₁). this gives us the gradient of (-5-3)/(8-6)=-8/2= -4. Since the gradient of the radius is -4, we know that the gradient of the tangent at this point is the negative inverse of this, so 1/4.
Using this gradient and the point K(8,-5), we can get the equation of the line using the formula y-b=m(x-a) y+5=1/4(x-8) Then multiply out the brackets and rearrange for yy+5=1/4x - 2y= 1/4x - 7 And that is the final answer