to circle x^2+y^2-4y-36=0

book has answers of

1) y=-3x+22

and

2) x-3y+26=0

how does one get there or is the book wrong?

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- Apr 22nd 2012, 05:09 AMsluggerbrothfind equation of lines thru (4,10) and tangent
to circle x^2+y^2-4y-36=0

book has answers of

1) y=-3x+22

and

2) x-3y+26=0

how does one get there or is the book wrong? - Apr 22nd 2012, 07:11 AMHallsofIvyRe: find equation of lines thru (4,10) and tangent
First, any (non-vertical) line can be written y= mx+ b. You are told that the line passes throught (4, 10) so when x= 4, y= 10. You have 10= m(4)+ b so b= 10- 4m and you can write the the equation of the line as y= mx+ 10- 4m.

In order to be tangent to the line must touch that circle at some point [tex](x_0, y_0)[/itex] and must have the same derivative there. The derivative of y= mx+ b is, of course, the slope, m, while the derivative of is given by 2x+ 2yy'- 4y'= 0 so y'= 2x/(2y- 4)= x/(y- 4). That is, you must have , , and , three equations for , , and .

That method will work for a tangent to any curve. Because this is specifically a circle, we can do this more simply using geometric properties. Again, let be the point of tangency. The equation of the circle is [tex]x^2+ y^2 -4y- 36= x^2+ y^2- 4y+ 4- 40= x^2+ (y- 2)^2- 40= 0 img.top {vertical-align:15%;} to (0, 2) is and the tangent line is perpendicular to that radius and so has slope .

That is, the line has equation . That, together with [itex]x_0^2+ (y_0- 2)^2= 40" alt=". That is, the circle has (0, 2) as center so the slope of the line from to (0, 2) is and the tangent line is perpendicular to that radius and so has slope .

That is, the line has equation . That, together with [itex]x_0^2+ (y_0- 2)^2= 40" /> gives**two**equations to solve for and and then you can put those into the equation of the tangent line.

In either case, you will be solving a quadratic equation which can have two solutions, giving two tangent lines.