Tangents to a Circle from a point not on the circle?

Question

Any one know how to do this

Find the equatiobs if the lines parallel to the line 3x+4y-6=0 and which are tangents to the circle x(squared) + y(squared) - 2x - 2y - 7 = 0

Answer 1

circle = x^2 + y^2 - 2*x - 2*y = 7 which simplifies to,

(x-1)^2 + (y-1)^2 = 3^2,

or a circle of radius 3 centered at the point (1,1).

Do a coordinate transformation to the point (1,1) so the circle is centered at 0. xx = (x-1), yy = (y-1).

Then xx^2 + yy^2 = 9, and take the derivative to get the slope,

yy' = -xx / sqrt (9 - xx^2).

You want that to equal -3/4 which is the slope of the line,
y = -3/4 *x + 1.5

So want,

-xx / sqrt (9-xx^2) = -3/4 which has the solutions
xx = 9/5,
xx = -9/5

Last 3 steps are easy:
(a) solve for 2 values of yy given xx = 9/5, xx = -9/5 and equation of circle xx^2 + yy^2 = 9 (yy = +-12/5)

(b) solve for 2 values of b given (9/5, 12/5) and (-9/5, -12/5) (from step (a) above) using equation yy = -3/4*xx + b

(c) Transform yy = 3/4*xx + b back to coordinates x and y using equations xx = x - 1 and yy = y - 1

Answer 2

x^2 + y^2 - 2x - 2y - 7 = 0.............i
(x^2 -2x) + (y^2 -2y) -7 =0
(x^2 -2x +1) + (y^2 -2y +1) =7 +1+1
(x -1)^2 + (y -1)^2 = 9=(3)^2
The center of the circle are (1,1) and radius=3
The line parallel to given line 3x+4y-6=0 will be
3x+4y-k=0 where k is constant.............ii
Because the length of perpendicular from the cent re of the circle to the tangent is always equal to radius.
{3(1)+4(1)-k}/sq rt(3^2+4^2)=(+/-)3
{3+4-k}/5=(+/-)3
Taking + sign
7-k=15
k =-8...............iii
Taking - sign
7-k=-15
k =22...............iv
Therefore equation of the two tangents parallel to given line to given circle is
3x+4y+8=0 ..............v
3x+4y-22=0 .............vi

Answer 3

The line parallel to the given line will have the same slope and will touch the circle at a point (x,y) that satisfies the equation of the circle. Easiest way is to draw both the line and circle, use a straitedge to transfer the same slope to a point on the circle, then plug into point slope form to get the equation of the new line.

Answer 4

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Answer 5

not good at maths