Guys please teach me how to solve this one. I want to learn. The question is find an equation of each of the two lines having slope -4/3 that are tangent to the circle x^2 + y^2 + 2x -8y - 8 = 0.
Find the two lines from a given slope that are tangent to a given circle
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The equation of any line having slope $-\dfrac43$ can be written as $\displaystyle y=-\dfrac43x+c\ \ \ \ (1)$
Method $\#1:$
We have the equation of the circle $x^2=y^2+2x-8y-8=0\iff \{x-(-1)\}^2+(y-4)^2=5^2$
The radius $(=5)$ must be equal to the perpendicular distance of the tangent from the center $(-1,4)$
Method $\#2:$
Now the replace $y$ with $\displaystyle-\dfrac43x+c$ in the equation of the given circle to form a Quadratic equation in $x$
Each value of $x$ represents the abscissa of the intersection which has one ordinate associated that cane be found using $(1)$.
For tangency, the intersections must coincide i.e., the roots of the resultant Quadratic equation in $x$ must be same i.e., the discriminant must be zero which will give us the required values of $c$
Note the equation of the circle can be written as $(x + 1)² + (y - 4)² = 5^2$. This means the centre of the circle $O \equiv (-1;4)$, and its radius measures $r=5$. Now it's easier to set the system and find the equation of the 2 lines: \begin{cases} (x + 1)² + (y - 4)² = 5^2\\ y=-\frac43 x +q\\ \end{cases} You know that in order to be tangent, $\Delta$ must equal zero, so \begin{cases} y=\sqrt{5^2-(x+1)^2}+4\\ y=-\frac43 x +q\\ \end{cases} $$\sqrt{5^2-(x+1)^2}+4=-\frac43 x +q\\\sqrt{5^2-(x+1)^2}=-\frac43 x +q-4\\5^2-(x+1)^2=16-8 q+q^2+\frac{32 x}{3}-\frac{8 q x}{3}+\frac{16 x^2}{9}\\\frac{16 x^2}{9}+x^2+2x+\frac{32 x}{3}-\frac{8 q x}{3}+q^2-8q-8=0\\x^2\left(\frac{25}{9}\right)+x\left(\frac{38-8q}{3}\right)+(q^2-8q-8)=0\\\Rightarrow \Delta=\left(\frac{38-8q}{3}\right)^2-4\cdot\frac{25}{9}(q^2-8q-8)=0\\\Rightarrow q=-\frac{17}{3} \lor q=11$$ As a conclusion the 2 lines have equation $$y=-\frac43 x +-\frac{17}{3}\qquad y=-\frac43 x +11$$ As a visual proof of the answer, please see image below: