Answer to Question #107891 in Analytic Geometry for maria

Question

Find the value of mif it is given that the line y=mxis 4 units from the point (4,-1)
Find the equation of the ciFind the equation of the tangent with the slope 3/4

to the circle x^2+y^2=13
which is tangent to the x at (6,0) and tangent to the y-axis

Answer 1

Given equation of the line is

$$y = mx$$

The perpendicular distance from the point $(4,-1)$ to the line $y = mx$ is $4$ units

$$\begin{vmatrix} \frac {mx_1 -y_1}{\sqrt {m^2+1}} \end{vmatrix} = 4$$

Here $(x_1,y_1) = (4, -1)$

$$\begin{vmatrix} \frac {m(4) -(-1)}{\sqrt {m^2+1}} \end{vmatrix} = 4$$

$$\begin{vmatrix} \frac {4m +1}{\sqrt {m^2+1}} \end{vmatrix} = 4$$

Squaring on both sides,

$(4m+1)^2 = 4(m^2+1)$

$$16m^2 + 1+ 8m = 4m^2 + 4$$

$$12m^2 + 8m - 3 = 0$$

$$m = \frac {-b±\sqrt{b^2 -4ac}}{2a} = \frac {-8±\sqrt{8^2 -4(12)(-3)}}{2(12)} \\ =\frac {-8±\sqrt{64 +144}}{24}$$

$$=\frac {-8±\sqrt{208}}{24} = \frac {-8± 4 \sqrt{13}}{24} = \frac {-2±\sqrt{13}}{6}$$

The equation of tangent with slope m to a circle $x^2 +y^2 = r^2$ is

$$y = mx ± r \sqrt {1+m^2}$$

Here slope = $m=\frac {3}{4} \\ r = \sqrt{13} \\ point = (x,y) =(6,0)$

$$y = \frac {3}{4} x± \sqrt {13} \sqrt {1+ \frac {9}{16}}$$

$$y = \frac {3}{4} x ± \frac {5}{4} \sqrt {13}$$