Question

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

• A. $\frac{2}{3}$
• B. $\frac{3}{2}$
• C. $\frac{7}{24}$
• D. $\frac{24}{7}$

Hint:

To find the slope of tangent lines from an external point to a circle, write the tangent line equation, substitute into the circle equation, and set the discriminant of the resulting quadratic to zero for tangency condition.

Answer 1

The Correct Option is C

Step 1: Equation of the circle
Given circle: $x^2 + y^2 = 9$ with center $O(0,0)$ and radius $r = 3$. Step 2: Equation of tangent line
Let the tangent line passing through $P(3,4)$ have slope $m$.
Equation of the line: $y - 4 = m(x - 3)$ or $y = mx - 3m + 4$. Step 3: Condition for tangency
Substitute $y$ in the circle equation: \[ x^2 + (mx - 3m + 4)^2 = 9. \] Expanding: \[ x^2 + m^2 x^2 - 2m x (3m - 4) + (3m - 4)^2 = 9, \] \[ (1 + m^2)x^2 - 2m(3m - 4) x + (3m - 4)^2 - 9 = 0. \] For the line to be tangent, the quadratic in $x$ has a single root, so the discriminant $D = 0$: \[ D = [ -2m(3m - 4) ]^2 - 4 (1 + m^2) [ (3m - 4)^2 - 9 ] = 0. \] Simplify: \[ 4 m^2 (3m - 4)^2 - 4 (1 + m^2) [ (3m - 4)^2 - 9 ] = 0, \] Divide both sides by 4: \[ m^2 (3m - 4)^2 = (1 + m^2) [ (3m - 4)^2 - 9 ]. \] Step 4: Solve for $m$
Expand and simplify to find possible values of $m$: \[ m^2 (3m - 4)^2 = (1 + m^2) ( (3m - 4)^2 - 9 ). \] Rearranged: \[ m^2 (3m - 4)^2 = (3m - 4)^2 + m^2 (3m - 4)^2 - 9 - 9 m^2, \] \[ 0 = (3m - 4)^2 - 9 - 9 m^2. \] Expanding $(3m -4)^2 = 9m^2 - 24m + 16$, \[ 0 = 9 m^2 - 24 m + 16 - 9 - 9 m^2, \] \[ 0 = -24 m + 7, \] \[ 24 m = 7, \] \[ m = \frac{7}{24}. \] Thus, the slope of the non-vertical tangent is $\frac{7}{24}$.