1. C2 Circle Co-ordinate Geometery

A circle C has centre D and equation
x² + y² + 2x – 8y + 8 = 0.
(a) Find the coordinates of D and the radius of C. [3]
(b) A line is drawn through the point P (4, 6) so that it touches the circle C at the point T.
(i) Show that PT = √20.
(ii) Find the equation of the circle centre P which passes through the point T. [5]

I know that a] will give D (-1, 4) and a radius of 3
But I have difficulty finding b and c. Can someone please help? I am having a lot of trouble with this and really need a bit of guidance.
Any help is appreciated

2. Originally Posted by db5vry
A circle C has centre D and equation
x² + y² + 2x – 8y + 8 = 0.
(a) Find the coordinates of D and the radius of C. [3]
(b) A line is drawn through the point P (4, 6) so that it touches the circle C at the point T.
(i) Show that PT = √20.
(ii) Find the equation of the circle centre P which passes through the point T. [5]

I know that a] will give D (-1, 4) and a radius of 3
But I have difficulty finding b and c. Can someone please help? I am having a lot of trouble with this and really need a bit of guidance.
Any help is appreciated
Because a tangent is perpendicular to the radius in the tangent point the points D, P, T are the vertices of a right triangle with PD as hypotenuse:

$|\overline{PD}|=\sqrt{(4-(-1))^2 + (6-a)^2} = \sqrt{29}$

One leg of the right triangle is the radius with a length of 3. Use Pythagorean theorem:

$|\overline{PT}|=\sqrt{29-3^2}=\sqrt{20}$

The general equation of the circle with $M(x_M, y_M)$ and radius r is:

$(x-x_M)^2 + (y-y_M)^2=r^2$

You know the coordinates of the centre and you know the length of the radius: $r = |\overline{PT}|$

Therefore the equation of the circle is:

$(x-4)^2+(y-6)^2=20$