Problems from Intersection of a circumference and a straight line

Study the relative position of the circumference $$x^2+y^2-2x-3=0$$ and the straight line $$3x+y-5=0$$.

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Development:

We have the following system of two equations: $$$\left\{\begin{array}{c} x^2+y^2-2x-3=0 \\ 3x+y-5=0 \end{array}\right\} \Rightarrow x^2+(5-3x)^2-2x-3=0 \Rightarrow 5x^2-16x+11=0$$$

We compute the discriminant of the equation of the second degree: $$$x=\dfrac{16\pm\sqrt{16^2-4\cdot5\cdot11}}{2\cdot5}=\dfrac{16\pm\sqrt{36}}{10} \Rightarrow \Delta=36 > 0$$$ The straight line and the circumference are secant since the discrimant is greater than zero.

Let's calculate the two intersection points. $$$x=\dfrac{16\pm\sqrt{36}}{10} \Rightarrow x=\left\{\begin{array}{c} \dfrac{11}{5} \\ 1 \end{array}\right. \Rightarrow y=\left\{\begin{array}{c} \dfrac{-8}{5} \\ 2 \end{array}\right.$$$

Solution:

They are secant at points of the circumference $$\Big(\dfrac{11}{5},-\dfrac{8}{5}\Big)$$ and $$(1,2)$$.

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Study the relative position of the circumference $$x^2+y^2-4x+2y-20=0$$ and the straight line $$3x+4y-27=0$$.

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Development:

We set out the system $$$\left\{\begin{array}{c} x^2+y^2-4x+2y-20=0 \\ 3x+4y-27=0 \end{array}\right.$$$

We isolate $$x$$ using the equation of the straight line: $$$x=\dfrac{27-4y}{3}$$$ and we replace it in the general equation of the circumference that we have $$$\Big(\dfrac{27-4y}{3}\Big)^2+y^2-4\Big(\dfrac{27-4y}{3}\Big)+2y-20=0$$$ It is the equation we were looking for. Now we need to compute the discrimant. Developing the squares we have: $$$\dfrac{27^2}{9}-2\cdot\dfrac{27}{3}\cdot\dfrac{4y}{3}+\dfrac{16y^2}{9}+y^2-\dfrac{4\cdot27}{3}+\dfrac{16y}{3}+2y-20=0$$$ $$$\Big(\dfrac{16}{9}+1\Big)y^2+\Big(\dfrac{16}{3}+\dfrac{6}{3}-\dfrac{72}{3}\Big)y-20+\dfrac{27^2}{9}-\dfrac{4\cdot27}{3}=0$$$ $$$\Big(\dfrac{25}{9}\Big)y^2+\Big(-\dfrac{50}{3}\Big)y-20+27\cdot3-\dfrac{108}{3}=0$$$ $$$\Big(\dfrac{25}{9}\Big)y^2+\Big(-\dfrac{50}{3}\Big)y+\dfrac{75}{3}=0$$$ $$$25y^2-150y+225=0$$$

Solving this equation of the second degree we have: $$$y=\dfrac{150\pm\sqrt{150^2-4\cdot25\cdot225} }{2\cdot25}=\dfrac{150\pm\sqrt{0}}{50}=3$$$ Therefore the discriminant is $$$\Delta=\sqrt{22.500-22.500}=0$$$ Thus, we can conclude that the circumference and the line are tangent at a point. We can put back the value $$y=0$$ to obtain the value of $$x$$: $$$x=\dfrac{27-4\cdot3}{3}=\dfrac{27-12}{3}=\dfrac{15}{3}=5$$$ therefore the intersection point will be $$(5,3)$$.

Solution:

They are tangent at point $$(5,3)$$.

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