# Which of the circle the line px+qy+r=0 will intersect?

If p, q and r are in arithmetic progression,
then the line

px + qy + r = 0


necessarily intersects which of the following circles?

$$x^2 + y^2 + 4x – 4y + 7 = 0$$ or $$x^2 + y^2 – 6x + 6y + 13 = 0$$

I tried by assigning

p=a

q= a+d

r =a+2d

where a = first term and d =common-difference.

I put these values in

px + qy = r

And put this value of x in terms of y in the circle equations. But then it get complicated. And i got stuck. Is there is any other easier way it can be solved ? Thanks in advance.

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The equation of the first circle is not quite clear. Needs to edited? – user21436 Mar 20 '12 at 7:36
It often simplifies arithmetic to write the arithmetic progression as $a-d , a, a+d$. Do you know how to find the distance of a point from a line? – Mark Bennet Mar 20 '12 at 7:49
@Thanks.I wil take care from next time.Ya..i know – vikiiii Mar 20 '12 at 7:52
In your final equation you need $px+qy=-r$ to be consistent with the line given by the first equation. – Mark Bennet Mar 20 '12 at 7:53

First, prove that regardless of the values of p, q and r, the line must pass through the point (1, -2). Next, find out which of the two circles has this point on the inside. You can then argue that the line must pass through this circle.

Post again if you need help with either of the two main steps, and I'll add more detail to this answer.

Edit

If p, q, r are in A.P., then $r=2q-p$. So write $px+qy+r=0$ as $p(x-1)+q(y+2)=0$. But $(1,-2)$ is always a solution to this, so this point must be on the line, regardless of $p$ and $q$.

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How do you get point (-1,2 ).And why it must pass through this? – vikiiii Mar 20 '12 at 7:53
OK, see my edit for that. I also noticed that the line has a different equation in the question title than in the body of the question. I answered the version from the question body, not the question title; let me know if that was the wrong one. – user22805 Mar 20 '12 at 8:02
i have edited the title. – vikiiii Mar 20 '12 at 8:10
Thanks.I understood after you added your edit.:) – vikiiii Mar 20 '12 at 13:04
So since $P=(1,-2)$ is on the second circle centered at $(3,-3)$ with radius $\sqrt5$, the line $p(x-1)+q(y+2)=0$ intersects the circle once if $p+2q=0$ and twice otherwise. And since $P$ is $5$ units from the other (unit) circle's center at $(-2,2)$, it intersects that circle iff $$\left|\tan^{-1}\frac{q}{p}+\tan^{-1}\frac{4}{3}\right|<\sin^{-1}\frac15,$$ once at each endpoint and twice at each interior point of the interval. – bgins Mar 21 '12 at 23:49

Rewrite equations of the circles into following forms :

$C_1 : (x+2)^2+(y-2)^2=1$

$\Rightarrow (x_O,y_O)=(-2,2) ~\text{and}~ R_1=1$

$C_2 : (x-3)^2+(y+3)^2=5$

$\Rightarrow (x_O,y_O)=(3,-3) ~\text{and}~ R_2=\sqrt 5$

Next , define distance of the center of circle from the line as :

$$d=\frac{|px_O+qy_O+r|}{\sqrt{p^2+q^2}}$$

Now , consider Intersection criterion :

$\begin{cases} \text{the line intersect a circle}, & \text{if }~ d<R\\ \text{the line is tangent of circle}, & \text{if }~d=R\\ \text{line and circle have no common points}, & \text{if }~d>R \end{cases}$

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$x^2 + y^2 + 4x – 4y + 7 = 0$ can be written as $(x+2)^2+(y-2)^2=1^2$ and $x^2 + y^2 – 6x + 6y + 13 = 0$ as $(x-3)^2+(y+3)^3 = {\sqrt{5}}^2$

The first circle has center $(-2, 2)$, radius $1$ and the second with center at $(3,-3)$ and radius $\sqrt{5}$.

These two circles do not intersect (why?)

Also your equation of line that has coefficients in A.P as David suggested can be written as $p(x-1)+q(y+2)=0$, which always passes through $(1,-2)$ and among the two circles only the second circle passes through $(1,-2)$

Work on showing this, and also showing that it does not intersect the first circle.

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