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enter image description here

My method

I know the the centre of the circle is at the point $(6,5)$ and that the radius is root $17$. Based on the diagram, I drew a line from the centre to the y axis at the given point $(0,12)$ and also extended two lines from the centre to the tangents to make two triangles.

I know that the line from the centre to the y axis is the hypothenuse of the two triangles, I worked out the distance between them and got root $85$ BUT I didn't know where to go from there.

I guess I could take the radius squared away from the hypothenuse to find the length of the tangent but then I don't see where I can find the other coordinates I need to ge the gradient to form the equations of the actual tangents.

I know this question is quite long but I will appreciate all the help. Thank you!

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  • $\begingroup$ Why the downvote, o anonymous voter? $\endgroup$ – amd Nov 4 '17 at 23:14
  • $\begingroup$ Continuing the way you’re going, once you find the distance from $(0,12)$ to the points of tangency, you could then find the intersection of the given circle with a circle centered at $(0,12)$ to find those points. $\endgroup$ – amd Nov 4 '17 at 23:17
  • $\begingroup$ Hi, thanks for your reply. I've found the distance of the tangents however I am still unsure on how I'd find the coordinates of intersections @amd. $\endgroup$ – Hanka Langerova Nov 4 '17 at 23:19
  • $\begingroup$ Do you know how to solve systems of equations? For a pair of circles, you can subtract one equation from the other to get a linear equation in $x$ and $y$. You can solve that for $x$ or $y$ in terms of the other variable and then substitute back into one of the circle equations to get a single-variable quadratic equation. $\endgroup$ – amd Nov 4 '17 at 23:22
  • $\begingroup$ Do you mean I should substiture either $x = 0$ or $y = 12$ into the equation of the circle to find the coordinates of intersection? $\endgroup$ – Hanka Langerova Nov 4 '17 at 23:25
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To continue the way you’ve started, you know that the distance from the circle’s center $C$ to the point $P=(0,12)$ is $\sqrt{85}$ and the radius of the circle is $\sqrt{17}$, so the distance from $P$ to the points of tangency is $\sqrt{85-17}=2\sqrt{17}$. The points of tangency are thus the intersections of the given circle with the circle $x^2+(y-12)^2=68$.

To solve this system of equations, start by expanding them into $$\begin{align} x^2+y^2-12x-10y+44 &= 0 \\ x^2+y^2-24y+76 &= 0 \end{align}$$ and then subtracting one from the other to eliminate the quadratic terms: $$12x-14y+32=0.$$ You can now solve this equation for $x$ or $y$ and then substitute back into one of the circle equations to get a quadratic equation in $x$ or $y$ that I expect you know how to solve. Since the second circle’s equation only has a single terms that involves $x$, I’ll solve for $x$ and substitute into that one: $$x = \frac76y-\frac83\tag1$$ and, after some manipulation, $${17\over36}(5y^2-64y+176)=0.\tag2$$ Solve equation (2) for $y$ and get the corresponding $x$ values from equation (1). You then have the two points of tangency and I hope that you can build the equations of the tangent lines from them.

Here is a different, much less involved way to solve this problem. You can tell from the diagram (and can prove mathematically) that neither tangent will be vertical, so the equations of the tangent lines can be written in the form $y=mx+12$. Substitute this into the equation of the circle and rearrange: $$(x-6)^2+(mx+12-5)^2 = 17 \\ (m^2+1)x^2+(14m-12)x+68 = 0.$$ For the line to be tangent to the circle, this equation must have a repeated real root, i.e., the discriminant $(14m-12)^2-4\cdot68(m^2+1)$ must be equal to zero. Solve this equation for $m$ and you have the equations of the tangent lines without further ado.

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