Problems from Intersection of a circumference and a straight line

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

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Development:

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

We compute the discriminant of the equation of the second degree: $x = \frac{16 \pm \sqrt{16^{2} - 4 \cdot 5 \cdot 11}}{2 \cdot 5} = \frac{16 \pm \sqrt{36}}{10} \Rightarrow Δ = 36 > 0$ The straight line and the circumference are secant since the discrimant is greater than zero.

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

Solution:

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

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

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Development:

We set out the system ${\begin{matrix} x^{2} + y^{2} - 4 x + 2 y - 20 = 0 \\ 3 x + 4 y - 27 = 0 \end{matrix}$

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

Solving this equation of the second degree we have: $y = \frac{150 \pm \sqrt{150^{2} - 4 \cdot 25 \cdot 225}}{2 \cdot 25} = \frac{150 \pm \sqrt{0}}{50} = 3$ Therefore the discriminant is $Δ = \sqrt{22.500 - 22.500} = 0$ 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 $x$ : $x = \frac{27 - 4 \cdot 3}{3} = \frac{27 - 12}{3} = \frac{15}{3} = 5$ therefore the intersection point will be $(5, 3)$ .

Solution:

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

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