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Question:

Find the center of family of circles cutting the family of circles: $$x^2+y^2+4x(\lambda-\frac{3}{2})+3y(\lambda - \frac{4}{3})-6(\lambda+2)=0$$ orthogonally.

My attempt:

Resolving it we get:

$$(x^2+y^2-6x-4y-12) + \lambda(4x+3y-6)=0$$ which is a family of circles passing through the points of intersection of the line and the circle. I'm having trouble doing the "cutting orthogonally" part. Can I get some hints there?

Thanks!

Update: This question was asked as an objective question (one question, four options, only one correct) and so deserves a short approach.

Up-to-date Update: These are the options (I won't reveal the correct answer because I don't want contrived answers)

(a) $x-y-1=0$ (b) $4x+3y-6=0$ (c) $4x+3y+7=0$ (d) $3x-4y-1=0$

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    $\begingroup$ Isn't any other info given? Something that'll allows to solve for $\;\lambda\;$ or stuff? Because as it is I get a rather ugly equation with too many unknowns...the circles in that family have center at $\;\left(\,-2\lambda+3,\,-\frac32\lambda+2\,\right)\;$ , so an orthogonal circle to these will fulfill that the sum of both circles' squared radius equals the squared distance from center to center...ugly. $\endgroup$
    – DonAntonio
    Commented Jan 29, 2017 at 8:50
  • $\begingroup$ @DonAntonio Indeed that's true. But no other info is given :( $\endgroup$
    – Gaurang Tandon
    Commented Jan 29, 2017 at 8:55
  • $\begingroup$ Then I'm afraid it is going to be awfully ugly and lengthy...unless there's some slicky trick (algebraic, geometric or whatever) that I missed. Sorry. $\endgroup$
    – DonAntonio
    Commented Jan 29, 2017 at 9:00
  • $\begingroup$ @GaurangTandon can you list the options? $\endgroup$
    – RSS
    Commented Jan 29, 2017 at 17:35
  • $\begingroup$ @foofaafuu Added please check. $\endgroup$
    – Gaurang Tandon
    Commented Jan 30, 2017 at 10:22

1 Answer 1

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Concerning families of circles, the following are well-known:

(1) If a circle cuts two circles of the family orthogonally, it cuts all elements of the family orthogonally.

(2) The circles which cut the circles in a family orthogonally form another family of circles, called the conjugate family of circles.

(3) The center of the conjugate family of circles is the radical line of the original family of circles.

Thus we need to find the radical line of our given family of circles. This is the radical line of any two circles in the family, i.e., the locus of points whose power with respect to $$(x^2+y^2−6x−4y−12)+\lambda_1(4x+3y−6)=0$$ and $$(x^2+y^2−6x−4y−12)+\lambda_2(4x+3y−6)=0$$ is equal for some given $\lambda_1, \lambda_2$. The power of $(x,y)$ with respect to the first circle is $(x^2+y^2−6x−4y−12)+\lambda_1(4x+3y−6)$ and wrt the second circle is $(x^2+y^2−6x−4y−12)+\lambda_2(4x+3y−6)$, thus our locus is $$(x^2+y^2−6x−4y−12)+\lambda_1(4x+3y−6)=(x^2+y^2−6x−4y−12)+\lambda_2(4x+3y−6),$$ $$4x+3y−6=0.$$

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