AIEEE 2009

Geometry Level 5

If $P$ and $Q$ are the points of intersection of the circles $x^2 + y^2 + 3x+7y + 2p-5 = 0$ and $x^2+y^2 + 2x+2y-p^2 = 0$ , then there is a cirlce passing thorugh $P,Q$ and the point $(1,1)$ for:

all except one value of

p

all except two values of

p

all values of

p

exactly one value of

p

4 solutions

Pranjal Jain
Dec 21, 2015

The equation of radical axis is $(x^2+y^2+3x+7y+2p-5)-(x^2+y^2+2x+2y-p^2)=0\\ \implies x+5y+p^2+2p-5=0$

To form a circumcircle, the line $PQ$ must not have $(1,1)$ on it. Thus, $p^2+2p+1\ne 0$ or $p\ne -1$ .

SHOULD WE CHECK CONDITION FOR A RADIUS IN FIRST CIRCLE, (3/2)^2+(7/2)^2-(2P-5)>0 . THIS GIVES . P<39/4.....

Amrit Anand - 5 years, 5 months ago

That would be unnecessary here.

Pulkit Gupta - 5 years, 5 months ago

Prakhar Bindal
Dec 21, 2015

Write the equation of circle passing through the point of intersection of two circles by using family of circles . and use the fact that arbitary constant cannot be -1 .

The general equation of the family of circles passing through the intersection of S1 and S2 in given by S1 + kS2 = 0, where k ≠ -1. Here again we have one-parameter (k) equation of family of circles. The particular value of the parameter gives a unique circle. if k = -1 then we will get radical axis of the two circles

Avi Ghanshani
Dec 21, 2015

The point should not lie on radical axis.

Amrit Anand
Dec 20, 2015

THIS ANSWER IS GIVEN BY ANSWER KEY. BUT I AM CONFUSED . SO POST YOUR SOLUTION SO THAT I CAN POST MY DAUBT............

I solved it using the concept of common chords. What's your doubt?

Pulkit Gupta - 5 years, 5 months ago

Can you elaborate on that? Thanks.

Pi Han Goh - 5 years, 5 months ago

Sure.

Define the first circle as $\large S_{1}$ & second circle as $\large S_{2}$ . Then, $\large S_{1}$ - $\large S_{2}$ shall give the equation of common chord of the two circles.

Let $\large S_{1}$ - $\large S_{2}$ = $\large L$ .The end points of this common chord $\large L$ ( the intersection points of two circles) are given to be P & Q in the question.

Equation of the circle passing through P & Q is given by $\large S_{1}$ + $\large k$ $\large L$ , where $\large k$ is any constant.

On solving, we obtain k as a fraction with ( $\large p$ - some constant ) as the denominator. Clearly, k is undefined for that constant value.

Pulkit Gupta - 5 years, 5 months ago

SHOULD WE CHECK CONDITION FOR A RADIUS IN FIRST CIRCLE, (3/2)^2+(7/2)^2-(2P-5)>0 . THIS GIVES . P<39/4.....

Amrit Anand - 5 years, 5 months ago

SHOULD WE CHECK CONDITION FOR A RADIUS IN FIRST CIRCLE, (3/2)^2+(7/2)^2-(2P-5)>0 . THIS GIVES .
P<39/4.....

Amrit Anand - 5 years, 5 months ago

AIEEE 2009

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