Problems with Circle question!?

Question

A circle has a centre (-1,4) and radius √12. A line is drawn through point P (4,6) so that it touches the circle C at point T. Show that PT=√20

Thank you!

Answer 1

This is a triangle problem. I get that PT = √17, not √20

Draw a line from point P(4,6) to the circle so it hits the circle at one point. Next draw a line from the center (C) to this same point creating a 90 degree angle.

The 3rd side of your triangle will be from the center of the circle to point P - this is the hypotnenuse of your triangle.

We can get that legth by using the distance formula
CP = √ ((4-(-1))^2 + (6-4)^2) = √(25 + 4) = √29

Now use pythagorus theorem
CT = √12
PT = ?
CP = √29

12 + PT^2 = 29
PT^2 = 17
PT = √17

Answer 2

P: (4,-6)
Eqn of circle: (x+1)^2 + (y - 4)^2 = 12
or x^2 +2x + 1 + y^2 - 8y + 16 = 12
x^2 + y^2 + 2x - 8y + 5 = 0
Eqn of line thru (4,-6) of slope s:
(y + 6)/(x - 4) = s
Let T (x, y) be the pt tangent to the circle
then
(1) ` ` ` (y + 6)/(x - 4) = s
(2) ` ` ` x^2 + y^2 + 2x - 8y + 5 = 0
and slope of circle at T = dy/dx =
2(x + 1) + 2(y - 4) dy/dx = 0
so
(3) ` ` ` dy/dx = s = - (x + 1) / (y - 4)
solve for (1), (2) & (3)
y + 6 = s(x - 4) = -(x - 1)(x - 4) / (y - 4)
(y + 6)(y - 4) = - (x - 1)(x - 4)
y^2 + 2y - 24 = - (x^2 - 5x + 4)
(4) ` ` ` x^2 + y^2 - 5x + 2y - 20 = 0
(2) ==> x^2 + y^2 = - (2x - 8y + 5)
- 2x + 8y - 5 - 5x + 2y - 20 = 0
- 7x + 10y = 25
keep solving for x and y ....

Answer 3

You get a right angled triangle formed by the center ,P and T(vertice of the right angle)
PC ^2= (6-4)^2+(4+1)^2 = 29
r^2=12 so
PT^2 = 29-12 =17 so PT = sqrt(17)
Check you data

Answer 4

centre O, T and P form a right angled tringle with right angle at T.
Using Pythagoras
OT^2+PT^2= OP^2
Using distance formula OP=sqrt((-1-4)^2+(4-6)^2)=sqrt 29
OT= rad =sqrt 12
Hence 12+PT^2=29
PT=sqrt17 and not sqrt 20.