The slope of the non-vertical tangent drawn f | vedclass

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(C) Let the slope of the tangent be $m$. The equation of the line passing through $(3, 4)$ with slope $m$ is $y - 4 = m(x - 3)$,which simplifies to $m

Vedclass · Accepted Answer

$\frac{7}{24}$

Vedclass · Answer

(C) Let the slope of the tangent be $m$. The equation of the line passing through $(3, 4)$ with slope $m$ is $y - 4 = m(x - 3)$,which simplifies to $mx - y + (4 - 3m) = 0$.Since this line is a tangent to the circle $x^2 + y^2 = 9$ (with center $(0, 0)$ and radius $r = 3$),the perpendicular distance from the center to the line must equal the radius.Using the formula $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$,we have $3 = \frac{|m(0) - 1(0) + 4 - 3m|}{\sqrt{m^2 + (-1)^2}}$.$3\sqrt{m^2 + 1} = |4 - 3m|$.Squaring both sides: $9(m^2 + 1) = (4 - 3m)^2$.$9m^2 + 9 = 16 - 24m + 9m^2$.$9 = 16 - 24m$.$24m = 7$.$m = \frac{7}{24}$.

The slope of the non-vertical tangent drawn from the point $(3, 4)$ to the circle $x^2 + y^2 = 9$ is

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