Find the number of values of $p$ for which the given condition is valid.

Question

If P and Q are 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 how many values of p?

From the family of circles eqaution

$$S_1+\lambda S_2=0$$ The circle passes through (1,1)

So $$(7+2p)+\lambda (6-p^2)=0$$ $$\lambda =\frac{7+2p}{p^2-6}$$

Then it should be valid for values of p except for $\sqrt 6$. But the answer says -1 is not valid. What is correct answer?

score 2 · Accepted Answer · answered Feb 15 '20 at 16:41

2

Hint

So, the equation becomes $$(1+\lambda)(x^2+y^2)+x(\cdots)+y(\cdots)+\cdots=0$$

What if $1+\lambda=0$

answered Feb 15 '20 at 16:41

lab bhattacharjee

274,582

But it wouldn’t be valid for $\pm \sqrt 6$ either, right? – Aditya Feb 15 '20 at 17:18
@Aditya, $$\lambda=?$$ if $$p^2=6$$ – lab bhattacharjee Feb 15 '20 at 18:22
I don’t understand – Aditya Feb 17 '20 at 12:27
@Aditya, First of all, why did you accept it, if not understood? What value will $\lambda$ have if $p^2=6$ – lab bhattacharjee Feb 17 '20 at 12:42
$\infty$, which is why the $\pm \sqrt 6$ shouldn’t be valid – Aditya Feb 17 '20 at 12:47
@Aditya, So, where is the confusion ? – lab bhattacharjee Feb 17 '20 at 12:50
That’s what I asked. The numbers for which the condition isn’t valid is $-1,\pm \sqrt 6$ but it just says -1 is the correct option – Aditya Feb 18 '20 at 11:34
@Aditya, Options may always not be correct. Have you identified the reduced conics when $$\lambda\to\infty?$$ – lab bhattacharjee Feb 18 '20 at 11:54
The answer below this, says that $\pm \sqrt 6$ is in fact valid, but I wasn’t convinced – Aditya Feb 18 '20 at 12:44
@Aditya, If $$\lambda\to\infty, S_2=-\dfrac{S_1}{\lambda}=0$$ right? – lab bhattacharjee Feb 18 '20 at 12:45
Yes, but how does that prove that $\pm \sqrt 6$ is not valid for the required condition? To prevent further miscommunication, let me explain myself. I wish to say that $-,\pm \sqrt 6$ are values for $\lambda$ for which there are no solutions. But the answer below this says that $\pm \sqrt 6$ do satisfy the required conditions. So what is the actual right answer? – Aditya Feb 18 '20 at 13:35
@Aditya, $$\lambda+1\ne0$$ – lab bhattacharjee Feb 18 '20 at 13:44
Yes, I understand that. I am convinced that -1 is the invalid case. But in my answer above $\frac{7+2p}{p^2-6}$, it gives $\pm \sqrt 6$. Have I made a computation error, or is it a conceptual problem, or is $\pm \sqrt 6$ does give an invalid answer? – Aditya Feb 18 '20 at 17:03

amd · Answer 2 · 2020-04-13T21:23:03.083

You’ve made a slightly subtle, but common error. Observe that your “family of circles equation” omits $S_2$ itself: there’s no value of $\lambda$ for which $S_1+\lambda S_2$ is a multiple of $S_2$. That’s going to be a problem if $S_2$ passes through the point $(1,1)$, which it does when $p^2=6$.

You should instead start with the affine combination $(1-\lambda)S_1+\lambda S_2$, which doesn’t suffer from this flaw. It does omit the combination $S_1-S_2$, but that’s OK because that combination produces a line instead of a circle, anyway. If you work through the same calculations as before, you’ll end up with $\lambda(p+1)^2 = 2p+7$, which gives you a unique $\lambda$ unless $p=-1$, in agreement with the given solution.

That aside, I disagree with the given solution. The two circles only intersect when $p$ is in the (approximate) intervals $[-8.469,-1.744]$ and $[0.876,5.336]$†. Outside of these intervals there are no intersection points $P$ and $Q$ in the first place, so unless you’re allowing for imaginary intersection points, I would say that there are many more values of $p$ besides $-1$ for which there is no circle through $P$, $Q$ and $(1,1)$. Even taking this into consideration, though, $p=\pm\sqrt6$ do lie within the region in which the circles intersect, so your solution is still incorrect.

† More precisely, the endpoints of the intervals are the roots of $p^4+4p^3-44p^2-44p+69$, which are the values of $p$ for which the circles are tangent.

Is it $\lambda S_1^2$ or $\lambda S_2^2$? – Aditya Feb 17 '20 at 12:27 — Aditya, Feb 17 '20 at 12:27
@Aditya Neither. It’s $\lambda S_2$. Sorry about that. – amd Feb 17 '20 at 17:32 — amd, Feb 17 '20 at 17:32

Find the number of values of $p$ for which the given condition is valid.

2 Answers2