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Prove $a+b+1=0$ given that lines $x=3$ and $y=-4$ are tangents to a circle with centre $(a,b)$ in 2nd quadrant, and another tangent touches the circle at $(1, -3)$.

I am not given the radius of this circle, and hence I cannot use the standard equation of a circle. I've thought of using the gradients of the tangents of the circle but I cannot see how this will help me. Any tip will be appreciated thanks!

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    $\begingroup$ We have $a<0, b>0$ For $$(x-a)^2+(y-b)^2=r^2$$ $x=3\implies(y-b)^2=r^2-(3-a)^2\implies r=\pm(3-a)$ Similarly $r=\pm(4+b)$ $\endgroup$ – lab bhattacharjee Jul 5 '17 at 9:57
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    $\begingroup$ $(a,b)$ is in second quadrant, hence $x=3$ is right to it, so we have $R=3-a$. Also, $y=-4$ is below it, hence $R=b+4$. We have $3-a=b+4\implies a+b+1=0$. $\endgroup$ – Galc127 Jul 5 '17 at 9:57

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