math

Find the equation of the tangent lines to x^2 + y^2-16x -12y +87 =0 at points where y = 4.

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\(x^2 + y^2-16x -12y +87 =0\\ (x-16)x+(y-6)y+87=0\\ (x-8)^2+(y-6)^2-13=0\\ (y-6)^2=13-(x-8)^2 \)

\(y=\pm \sqrt{13-(x-8)^2}+6 \\ \large \color{blue}y=-\sqrt{(\sqrt{13})^2-(x-8)^2}+6\\\)

y = 4

\(\sqrt{13-(x-8)^2}+6=4\\ \sqrt{13-(x-8)^2}=-2\\ 13-(x-8)^2=4\\ x-8=\pm \sqrt9\\ \)

x ∈ { 5, 11 }

\(\large \color{blue}P_{f(x)}(5,\ 4)\\\large \color{blue} P_{g(x)}(11,\ 4)\)

\(y=- { (13-(x-8)^2})^{0.5}+6\\ \frac{dy}{dx}= -0.5\cdot (13-(x-8)^2)^{-0.5}\cdot 2\cdot (x-8)\\ \large \color{blue}\frac{dy}{dx}= \frac{x-8}{ \pm \sqrt{ 13-(x-8)^2}}\\ \)

\(m_{f(5)}=\frac{5-8}{+ \sqrt{ 13-(5-8)^2}} \\ \large \color{blue}m_{f(5)}= -1.5\\ m_{g(11)}= \frac{11-8}{ -\sqrt{ 13-(11-8)^2}}\\ \large \color{blue}m_{g(11)}=1.5\)

\(f(x)=m_f(x-x_f)+y_f\\ f(x)=-1.5(x-5)+4\\ \large \color{blue}f(x)=-1.5x+11.5\\ g(x)=m_g(x-x_g)+y_g\\ g(x)=1.5(x-11)+4\\ \large \color{blue}g(x)=1.5x-12.5\)

\(The\ radius\ of\ the\ circle\ is\ \sqrt{3}.\)

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