The circle has equation $(x-6)^2+(y-5)^2=17$. The lines $L_1$ and $L_2$ are each a tangent to the circle and intersect at the point $(0,12)$. Find the equation of $L_1$ and $L_2$ giving your answer in the form $y=my+c$.
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See https://math.stackexchange.com/questions/2254073/the-line-y-mxc-is-a-tangent-to-x2y2-a2-if – lab bhattacharjee Oct 26 '17 at 07:10
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Any line through the point $(0,12)$ which is not vertical has equation $y=mx+12$ with $m\in\mathbb{R}$. A line is tangent to the circle $(x-6)^2+(y-5)^2=17$ if and only if it intersects the circle at just one point. That is, if and only if, the following quadratic equation in $x$ has just one root $$(x-6)^2+((mx+12)-5)^2=17.$$ Can you take it from here? Give it a try and show your work!
Robert Z
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@Lynette Please re-edit your question and, below it, show your work so far. Let's see what it going wrong. – Robert Z Oct 26 '17 at 08:34
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Simplified to x^2+m^2x^2-12x+14mx+38=0 using the fact that the discriminate must be 0 – Lynette Oct 26 '17 at 08:36
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Thanks Robert. I can’t send the picture but you could sketch it. Sketch a circle at (6,5) and draw two tangents to the circle. These two tangents intersect at (0,12) – Lynette Oct 26 '17 at 08:55
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Well done. Now evaluate the discriminant. As you said, it has to be zero. – Robert Z Oct 26 '17 at 09:01
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@Lynette Glad to be useful. BTW, if you are new here, welcome to MSE and please take a few seconds for a tour https://math.stackexchange.com/tour – Robert Z Oct 26 '17 at 09:51