Finding the locus of the centre of the circle which is orthogonal to a family of circles

Question

The following question is from the Chapter "Circles":

Find the locus of the centre of family of circles cutting the family of circles represented by $x^2+y^2+4x(\lambda-\frac 3 2)+3y(\lambda-\frac 4 3)-6(\lambda+2)=0$ orthogonally.

My Approach:

The given equation representing the family of circles,

$x^2+y^2+4x(\lambda-\frac 3 2)+3y(\lambda-\frac 4 3)-6(\lambda+2)=0$

can also be represented by the following equation, where I've separated the $\lambda$ from rest of the equation

$(x^2+y^2-6x-4y-12)+\lambda (4x+3y-6)=0$

The above equation represents the family of circles passing through the intersection points of the circle $x^2+y^2-6x-4y-12=0$ and the line $4x+3y-6=0$.

I know that for two given circles : $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ to be orthogonal, the condition $2g_1g_2+2f_1f_2=c_1+c_2$ must be satisfied. I think after getting an expression of this kind, it will be easier to compute the locus of the new family of circles. But, I face trouble in finding the appropriate condition, since there is no fixed circle given, and only the equation for family of circles is given.

Kindly explain how to solve this problem. Is there any other alternative approach to this?

Answer 1

The locus is the radical axis of the given family, and the radical axis is the line passing through the intersection points of

$$(x^2+y^2-6x-4y-12)+\lambda_1 (4x+3y-6)=0\tag1$$

and

$$(x^2+y^2-6x-4y-12)+\lambda_2 (4x+3y-6)=0\tag2$$

where $\lambda_1\not=\lambda_2$.

$(1)-(2)$ implies that the locus is $$\color{red}{4x+3y-6=0}$$

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Added : Let me add some explanations.

(3) As you noticed, each circle of the given family $$(x^2+y^2-6x-4y-12)+\lambda (4x+3y-6)=0$$ passes through the intersection points of the circle $x^2+y^2-6x-4y-12=0$ and the line $4x+3y-6=0$.

(4) By the way, if two circles cross, their radical axis is the line passing through their two crossing points.

(5) The center of any circle that cuts the two circles orthogonally must lie on the radical axis.

It follows from these that the locus we want is the radical axis of any two circles of the given family. The radical axis is the line passing through the intersection point of any two circles of the family. So, it immediately follows that the locus is the line $4x+3y-6=0$.

Answer 2

We need to stop at an intermediate step in finding the singular solution of a family of curves by the Clairaut C-discriminant method. The singular solution is the common point of concurrence. The procedure is rather simple.

$$(x^2+y^2-6x-4y-12)+\lambda ( 4x+3y-6)=0\tag1$$

is partially differentiated with respect to $\lambda$ and in the result what remains directly furnishes the required locus of circle centers. $$ 4x+3y-6=0 \tag2$$

Done.

EDIT1:

It is the radical axis of the circle family. The given case is for real intersections. We can have the radical axis not only with constant intercepted lengths like here but also equal tangent lengths.