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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 circle passing through P, Q, and (1, 1) for ...

a)all except 2 values of p. b)all values of p. c) all except 1 value of p.

My working: $$s2= x^2 + y^2 + 3x + 7y +2p – 5= 0$$ and $$s1=x^2 + y^2 +2x + 2y – p^2 = 0$$ Equation of required circle is $S1+kS2=0$ (family of circles) Putting (1,1), I got $$k= \frac{2p+7}{p^2-6}.$$

Therefore, p is not equal to +/- root 6, i.e. excepting two values of p.

However, by using another method, I'm getting a different answer. $$s2=x^2 + y^2 + 3x + 7y +2p – 5$$ and $$s1=x^2 + y^2 +2x + 2y – p^2 = 0$$ Equation of $PQ= S1-S2.$

Equation of line$ L(PQ)=x+5y+p^2+2p-5.$ Required equation= $S1+ kL=0.$

$$ k= \frac{-2p-7}{(p+1)^2}.$$ Therefore, p is not equal to -1, i.e. excepting one value of p.

DanielWainfleet
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Tapi
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  • So, is your question why you’re getting different answers? – amd Apr 20 '19 at 23:02
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    The strictly correct answer is “none of the above:” when $p=39/4$ the first “circle” is a single point and for $p\gt39/4$ its equation has no real solutions at all. – amd Apr 20 '19 at 23:10

1 Answers1

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The discrepancy arises from the way that you’ve parameterized the family of circles in your first approach. There’s no value of $k$ for which the combination $S_1+kS_2$ produces $S_2$ itself, which is fine as long as $S_2$ doesn’t pass through $(1,1)$, but it in fact does so when $p=\pm\sqrt6$. That’s the true significance of the values of $p$ that you found there.

You might try the combination $kS_1+S_2$ instead, but that suffers from the same problem: $S_1$ passes through $(1,1)$ when $p=-7/2$. You need to use the affine combination $(1-k)S_1+kS_2$ to capture all of the possibilities. This is in fact the same as your second method: $S_1+kL = S_1+k(S_2-S_1)=(1-k)S_1+kS_2$. (Contrary to what you wrote, you actually used $S_2-S_1$, not $S_1-S_2$.) Note that this set of affine combination doesn’t include $S_1-S_2$, but that’s a line, so we’re not interested in it.

It’s worth examining the case $p=-1$ to see why it’s special. For that value of $p$, the radical axis $L$ has the equation $x+5y=6$, which means that it passes through $(1,1)$. There is a linear combination of the two circles that passes through $(1,1)$, but it’s not a circle (unless you consider the line to be a circle with infinite radius).

amd
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  • For p=-1, the equation becomes x+5y=6 which passes through (1,1) leading to a case where P, Q and (1,1) are collinear, hence a circle is not possible. When p=-√6, x+5y=-1+2√6 which does not pass through (1,1). When p=√6, x+5y=-1-2√6 which also doesn't pass through (1,1). Also, for p=-7/2, L is 4x+20y+1=0 which doesn't pass through (1,1). Does this necessarily mean that the only exception is -1? Does that essentially mean that a family of straight lines and a circle is a more efficient method of tackling these kinds of problems? – Tapi Apr 21 '19 at 06:55
  • I don't think we need to consider the values of p where S1 and S2 cease to circles because it is already given that they are circles. Also, x^2+y^2+2x+2y–p^2=0 is always a circle. – Tapi Apr 21 '19 at 07:22
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    @LenaDas $S_1+kL$ is a family of circles, not lines. The issue is not one of efficiency, but of correctness. When you use $S_1+kS_2$, you omit $S_2$ itself, which is fine if you know that it can’t be a solution, but in this case it can be. To be absolutely sure you don’t miss anything, you need to use the affine combination $(1-k)S_1+kS_2 = S_1+k(S_2-S_1)=S_1+kL$. – amd Apr 21 '19 at 08:25
  • Incidentally, Plücker’s mu provides a way to write a linear combination of $S_1$ and $S_2$ that passes through a given point $P$ directly: $S_2(P)S_1-S_1(P)S_2$. If you apply this to the problem at hand, you’ll find that the quadratic terms vanish when $p^2+2p+1=0$, i.e., when $p=-1$ as found by other methods. – amd Apr 21 '19 at 22:36