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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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    In your final equation you need $px+qy=-r$ to be consistent with the line given by the first equation.2012-03-20

3 Answers 3

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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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    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.2012-03-21
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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 :

$\begin{cases} \text{the line intersect a circle}, & \text{if }~ dR \end{cases}$

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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.