Question

If the locus of mid pt. of chord of the circle
x²+y² - 2x +4y-1=0 is another circle, find the
centre of that circle given the chord passes
through (
-7,5).
​

Answer 1

There is a circle x2+y2=a2. On any line that cuts the circle in two distinct points(it is a secant), the points of intersection with circle are taken and at those two points I draw the tangents that intersect at some point, say (α,β). It's given that the tangents intersect at right angles at the point (α,β). I need to find the locus of the midpoint of the chord.

THE BOOK'S WAY:

Let the midpoint be (h,k).

Using T=S1 for the equation of chord, the chord is hx+ky=h2+k2

If the chord intersects the circle at (x1,y1) and (x2,y2), then by condition of perpendicularity

m1m2=−1

(

y1−β

x1−α

)(

y2−β

x2−α

)=−1

(x1x2+y1y2)+(α2+β2)=α(x1+x2)+β(y1+y2)

Then using the equation of chord and separately eliminating x,y we obtain quadratics

λx2−2λhx+λ2−a2k2=0

λy2−2λky+λ2−a2h2=0

where λ=h2+k2. Using the values of product of roots and sum of roots, the locus is

x2+y2−αx−βy+

1

2

(α2+β2−a2)=0

MY WAY:

I translate the origin to (α,β) so that x=X+α, y=Y+β

Circle becomes X2+Y2+2αX+2βY+α2+β2−a2=0

To get the equation of pair of lines that join the circle and chord intersection points to origin(translated), I homogenise the equation of circle, then retranslate the axes to previous origin by using X=x−α, Y=y−β and then put the sum of coefficent of x2 and coefficient of y2 equal to zero. Then I get an equation that doesn't match the book's answer. What's the flaw? Thanks in advance[Sorry if that's a long question]