Given, two lines are tangent to the circle, and they intersect at a point $(0,14)$ not on the circle, find the product of the slopes $m_1*m_2$ $$l_{1}: y = m_{1}x + c_{1}$$ $$l_{2}: y = m_{2}x + c_{2}$$ $$Circle: y^2 + x^2 = 49$$
I find the derivative of slope from the circle equation $$\frac{dy}{dx} = -x$$ For $x_{1}$ is a point that lies on the circle $$m_{1} = f'(x_{1}) $$
I create a point-slope equation using the point (0,14) $$y-14=(x+0)m_{1}$$ sub $m_{1} = -x$ $$y=-x^2+14$$
So here's where I'm not too sure I'm correct about. I try to find $x_{1}$ that lies on the circle by subbing $y = \sqrt{x^2+14}$ derived from the circle equation. $$\sqrt{x^2+14} = -x^2+14$$ $$0=x^4-29x^2+182$$ solve for x $$x=\frac{29+\sqrt{113}}{2} or \frac{29-\sqrt{113}}{2}$$ However, $x > radius$ for both results, which is not possible since it needs to lie on the circle. How do I proceed?
Answer sheet states m1*m2=-3
The equation of any tangent at $(7\cos t,7\sin t)$
$$x\cos t+y\sin t-7=0\iff y=7\csc t-x\cot t$$
So, the gradient is $=-\cot t$
If it passes through $(0,14)$
$$14\sin t=7\iff\sin t=\dfrac12\implies \cot t=\pm\sqrt{\csc^2t-1}=\pm\sqrt3$$
So, the product of the gradients $=(-\sqrt3)(\sqrt3)=?$