circles part b ..help !

[maths_arghh234]

hey ...please help with part (b) of this question ...

A circle C has the centre D and equation
x^2+y^2+2x-8y +8=0

a) find the coordinates of D and the radius of C centre = (-1,4) radius = 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= the square root of 20
(ii) Find the equation of the circle centre P which passes through the point T :?

thanks :!:

 

[masters]

hey ...please help with part (b) of this question ...

A circle C has the centre D and equation
x^2+y^2+2x-8y +8=0

a) find the coordinates of D and the radius of C centre = (-1,4) radius = 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= the square root of 20
(ii) Find the equation of the circle centre P which passes through the point T :?

thanks :!:

Hi maths_arghh234,

You found the center and radius correctly.

PT will be a tangent to the circle at T, and the radius CT is perpendicular to the tangent at the point of tangency T. Draw a segment from C to P to complete the right triangle.

Next, find the distance from C(-1, 4) to P(4, 6). This will be the hypotenuse of your right triangle.

\(\displaystyle CP=\sqrt{(4-(-1))^2+(6-4)^2}\)

A little algebra will reveal that \(\displaystyle CP=\sqrt{29}\)

Now you have a leg 3 units long (your radius to the point of tangency) and the hypotenuse of your right triangle \(\displaystyle \sqrt{29}\). Time for Mr. Pythagorus.

\(\displaystyle (\sqrt{29})^2=(TP)^2+3^2\)

Again, a little algebra will show that \(\displaystyle TP=\sqrt{20}\)

To find the equation of the circle with center P(4, 6) and passing through T, we will note that the radius will be the length TP which we found to be \(\displaystyle \sqrt{20}\)

Finally, put it all back into the general equation of a circle.

\(\displaystyle (x-h)^2+(y-k)^2=r^2 \Longrightarrow \boxed{(x-4)^2+(y-6)^2=20}\)

 

[maths_arghh234]

oohh brilliant thanks !!