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An line defined by $ax+by+c=0$ goes through the center of the circle $x^2+y^2+2x-4y+1=0$ and the same line is tangent to the circle $x^2+y^2-2x=3/5$. What are the values of $a,b$ and $c$? Can any body give me a hint that how to approach this

CardioidAss22
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Robhido 7
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  • Please describe any attempt you have made to solve this problem yourself – MathTrain Aug 27 '19 at 08:26
  • The center is $A=(-1,2)$. By Joachimthal's notation $s_1^2=s_{11}s$ or $(x x(A)+y y(A)-(x(A)+x)-3/5)^2=(x(A)^2+y(A)^2-2x(A)-3/5)(x^2+y^2-2x-3/5))$ which factors as $-4 ; \left(x + 3 ; y - 5 \right) ; \frac{3 ; x + y + 1}{5}$ defines the tangent pair. – Jan-Magnus Økland Aug 27 '19 at 09:03
  • @Jan-MagnusØkland That should be an answer instead of a comment. – amd Aug 28 '19 at 10:09
  • Once you’ve found the center of the first circle, this becomes a duplicate of https://math.stackexchange.com/q/219298/265466 and many other similar questions here. Note, by the way, that there are two distinct such lines, but for each there is in infinite number of values for $a$, $b$ and $c$. – amd Aug 28 '19 at 20:04

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The equation $ax+by+c=0$ is equivalent (without the case of a vertical line) to: $y=mx+q$. By the tangency conditions, I have: $$\left\{\begin{matrix} (x-1)^2+y^2=\frac8{5} \\ y=mx+q \end{matrix}\right.$$ Substituing $y=mx+q$ in the first equation, I obtain: $x^2(m^2+1)+2x(mq-1)+q^2+\frac3{5}=0$. Because the circle and the line have to be tangent, the discriminant has to be equal to $0$, so: $$4(m^2q^2-2mq+1)-4(m^2+1)(q^2-\frac3{5})=0$$ To solve this, I can use the condition on the line $y=mx+q$, in fact it's the same as: $$\left\{\begin{matrix} \frac{12}{5}m^2-8mq-4q^2+\frac{32}{5}=0 \\ q=m+2 \end{matrix}\right.$$ The first equation becomes: $$\frac{12}{5}m^2-8m^2-16m-4m^2-16m-16+\frac{32}{5}=0$$ or in other words: $$48m^2+160m+48=0$$ Taking the discriminant, I obtain: $$m=\frac{-160\pm\sqrt{160^2-4\cdot48^2}}{2\cdot48}=-3 \lor-\frac{1}{3}$$ From this: $q=-1$ and $q=\frac53$. The two lines are: $$y=-3x-1 \rightarrow y+3x+1=0$$ $$y=-\frac13x+\frac53\rightarrow3y+x-5=0$$ There are only two tangent lines from a point to a circle, so there aren't any other lines in the form $x=k$.

Matteo
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