Answer to Question #107891 in Analytic Geometry for maria

Question #107891
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
1
Expert's answer
2020-04-06T17:54:25-0400

Given equation of the line is

y=mxy = mx

The perpendicular distance from the point (4,1)(4,-1) to the line y=mxy = mx is 44 units


mx1y1m2+1=4\begin{vmatrix} \frac {mx_1 -y_1}{\sqrt {m^2+1}} \end{vmatrix} = 4

Here (x1,y1)=(4,1)(x_1,y_1) = (4, -1)


m(4)(1)m2+1=4\begin{vmatrix} \frac {m(4) -(-1)}{\sqrt {m^2+1}} \end{vmatrix} = 4


4m+1m2+1=4\begin{vmatrix} \frac {4m +1}{\sqrt {m^2+1}} \end{vmatrix} = 4

Squaring on both sides,

(4m+1)2=4(m2+1)(4m+1)^2 = 4(m^2+1)

16m2+1+8m=4m2+416m^2 + 1+ 8m = 4m^2 + 4

12m2+8m3=012m^2 + 8m - 3 = 0

m=b±b24ac2a=8±824(12)(3)2(12)=8±64+14424m = \frac {-b±\sqrt{b^2 -4ac}}{2a} = \frac {-8±\sqrt{8^2 -4(12)(-3)}}{2(12)} \\ =\frac {-8±\sqrt{64 +144}}{24}


=8±20824=8±41324=2±136=\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 x2+y2=r2x^2 +y^2 = r^2 is

y=mx±r1+m2y = mx ± r \sqrt {1+m^2}

Here slope = m=34r=13point=(x,y)=(6,0)m=\frac {3}{4} \\ r = \sqrt{13} \\ point = (x,y) =(6,0)


y=34x±131+916y = \frac {3}{4} x± \sqrt {13} \sqrt {1+ \frac {9}{16}}


y=34x±5413y = \frac {3}{4} x ± \frac {5}{4} \sqrt {13}


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