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 = mx"

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


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

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


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


"\\begin{vmatrix}\n \\frac {4m +1}{\\sqrt {m^2+1}} \n\\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\u00b1\\sqrt{b^2 -4ac}}{2a} = \\frac {-8\u00b1\\sqrt{8^2 -4(12)(-3)}}{2(12)} \\\\\n=\\frac {-8\u00b1\\sqrt{64 +144}}{24}"


"=\\frac {-8\u00b1\\sqrt{208}}{24} = \\frac {-8\u00b1 4 \\sqrt{13}}{24} = \\frac {-2\u00b1\\sqrt{13}}{6}"

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

"y = mx \u00b1 r \\sqrt {1+m^2}"

Here slope = "m=\\frac {3}{4} \\\\\nr = \\sqrt{13} \\\\\npoint = (x,y) =(6,0)"


"y = \\frac {3}{4} x\u00b1 \\sqrt {13} \\sqrt {1+ \\frac {9}{16}}"


"y = \\frac {3}{4} x \u00b1 \\frac {5}{4} \\sqrt {13}"


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