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Math Help - find equation of lines thru (4,10) and tangent

  1. #1
    Member sluggerbroth's Avatar
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    find 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?
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  2. #2
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    Re: 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 x^2+ y^2- 4y- 36= 0 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 x^2+ y^2- 4y- 36= 0 is given by 2x+ 2yy'- 4y'= 0 so y'= 2x/(2y- 4)= x/(y- 4). That is, you must have x_0^2+ y_0^2- 4y_0- 36= 0, y_0= mx_0+ 10- 4m, and m= x_0/(y_0- 4), three equations for x_0, y_0, and m.

    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 (x_0, y_0) 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%;} (x_0, y_0) to (0, 2) is \frac{y_0- 2}{x_0} and the tangent line is perpendicular to that radius and so has slope m= \frac{x_0}{2- y_0}.
    That is, the line has equation y= \frac{x_0}{2- y_0}x+ 10- \frac{4x_0}{2- y_0}. 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 (x_0, y_0) to (0, 2) is \frac{y_0- 2}{x_0} and the tangent line is perpendicular to that radius and so has slope m= \frac{x_0}{2- y_0}.
    That is, the line has equation y= \frac{x_0}{2- y_0}x+ 10- \frac{4x_0}{2- y_0}. That, together with [itex]x_0^2+ (y_0- 2)^2= 40" /> gives two equations to solve for x_0 and y_0 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.
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