Determine p so that the line q does not have any point in common with the circle

Question

I am preparing to take an entrance exam for a university in my country, which will happen soon. As part of my preparation, I have been practicing with some sample math tests provided by the university. However, I recently came across a math problem that I could not understand at all.

This is the exercise:

Determine the value of the parameter $p$ so that the line $q$ does not have any point in common with the circle. $$q : px + y − 1 = 0$$ and $$k : x^2 − 4x + y^2 − 6y − 3 = 0$$

(a) There are infinitely many such $p$. (b) $p \in (−\infty, 3) \cup (7, \infty)$ (c) None of the other options is correct. (d) $p = 7$ (e) There is no such $p$.

the correct solution is (e) There is no such $p$, but I don't know how to get there. My procedure was the next:

1- From $q : px + y − 1 = 0$, I isolated the $y$.

$y= -px +1$

2- I substituted $y$ into the equation of the circle

$$x^2 − 4x + (px -1)^2 − 6(-px +1) − 3 = 0$$

Getting:

$$x^2 -4x -10 + p^2x^2 +4px =0$$

And here I am lost. Does anyone know what the procedure is? I apologize if there is a misspelling, English is not my mother language

Answer 1

Hint

If $\ a, b, c\ $ are real numbers, then the equation $$ ax^2+bx+c=0 $$ has no real solution if and only if its discriminant, $\ \sqrt{b^2-4ac}\ $, is negative. Therefore, your line $\ q\ $ will have no point in common with your circle $\ k\ $ if and only if the discriminant of your equation $$ \big(1+p^2)x^2+4(p-1)x-\color{red}{8}=0 $$ is negative (I believe the value of the constant term in this equation should be $\ {-}8\ $, not $\ {-}10\ $). What is the discriminant of this equation? Are there any values of $\ p\ $ for which it is negative?

Answer 2

Note that $x^2-4x-8+p^2x^2+4px=0$ is a quadratic equation in x. Solving this equation for x will give us the x coordinates of points where the line and circle intersect.

To meet the required condition, there should be no such point. Or, the equation should have no solution(s). Also, if a quadratic has no real solution, its discriminant must be negative.

Here, $a=1+p^2, b= 4(p-1), c= -8$ for the given quadratic.

Answer 3

Rewrite the equation of the circle, by completing the squares: $$x^2-4x+4-4+y^2-6y+9-9-3=0\\(x-2)^2+(y-3)^2=4^2$$ This is the circle of radius $4$, centered on $(2,3)$

Note that $(0,1)$ is the intersection of all the lines $q$, independent of value of $p$. The distance from this point to the center of the circle is $\sqrt{(-2)^2+(-2)^2}=2\sqrt 2<4$. So there is at least one point on each line inside the circle. That means that any line will intersect the circle.