Find the radius and the centre of the circle?

Question

x^2 + y^2 = 6x - 8y

find the co-ordinates of the 2 points P,Q where the line y-2x-20 cuts the circle, the equations of the two tangents to the circle P,Q and the coordinates of the point where these two tangents meet.

plz can sum1 help me with this question ive been struggling with it for the past 3 hours. thanks

Answer 1

Mathemat... has shown you how to find circle's center and radius. Here's how to solve rest of question.

x^2 + y^2 = 6x - 8y <-- Equation #1

y = 2x+20 <-- Equation #2

So x^2 +(2x+20)^2 = 6x -8(2x+20)

x^2 + 4x^2 +80x +400 = 6x -16x -120

5x^2 +90x +520 = 0

x = [-90 +/- 50sqrt(3)]/10 = -9 +/- 5sqrt(3)

x = -9+5sqrt(3) and x = -9 -5sqrt(3)

Plug the two values of x into Equation #2 to find corresponding y values

This gives you the two points P(-9+5sqrt(3) , 2+10sqrt(3)) and

Q(-9-5sqrt(3), 2-10sqrt(3)).

If you differentiate Equation #1 implicitly, you get:

2x +2yy' = 6 -8y'

2yy' +8y' = 6-2x

y'(2y+8) = 6-2x

y' = dy/dx = (6-2x)/(2y+8)

You can now find the slope at P and at Q by plugging the coordinates of P and then Q into the above equation for dy/dx. You can then find the equation of each tangent using y = mx+b where m is the slope.

Now that you have the equations for the two tangents, you can solve them simultaneously to find their point of intersection.

It's a little messy what with the sqrt(3) involved, but you should be able to do it.

Answer 2

Use Completing the Square to put the equation in the form

(x - h)^2 + (y - k)^2 = r^2

where

(h, k) is the center

and r is the radius

x^2 + y^2 = 6x - 8y

x^2 - 6x + y^2 + 8y = 0

(x^2 - 6x + 9) + (y^2 + 8y + 16) = 9 + 16

(x - 3)^2 + (y + 4)^2 = 25

Center is (3, -4), radius is 5

Answer 3

x² - 6x + y² + 8y = 0

(x² - 6x + 9) + (y² + 8y + 16) = 25

(x - 3)² + (y + 4)² = 5²

Centre (3 , - 4)

radius = 5

Can`t do next part because you have made an error in presenting the equation of the line.

Sorry but not prepared to guess.

Answer 4

pye 22 over 7

Answer 5

i passed algebra already but i can't help u with that sorry.