Question

Show that the circle on the chord xcosα+ ysinα= p
Of the circle x^2+ y^2= a^2 as diameter is x^2+ y^2 -a^2-2p( xcosα+ ysinα- p)=0

Answer 1

If S ≣ x² + y² + 2gx + 2fy + c = 0 is a circle intersecting the straight line μ ≣ ax + by + c = 0 at two real and distinct points, then S + λμ = 0 is the equation of the family of circles passing through the points of intersection of S = 0 and μ = 0.
Here, S = $x^2+y^2-a^2=0$............(1)
The equation of the line (chord ) = x cos α +y sinβ =p , μ =x cos α +y sinβ -p =0...........(2)
so the circle having the chord as the diameter will be passing through the point of itersections of the given circel and its chord .
S + λμ = 0 gives:
($x^2+y^2-a^2$) + λ(x cos α +y sinβ -p ) =0...............(3)
Rearrange the equation in general form ,
x² + y² + λx cos α +λ y sin α -a² -λ p =0
Here , the center of the circle is (h,k) where h= -coeff of x /2 and k= - coeff of y/2)
h=-λ cosα /2 and k =λ sin α /2
Since the center will satisfy the equation of the diameter , we plug in the values of (x,y ) in the equation of the chord and solve for λ .
x cos α +y sinβ = p ..........equation of the diameter of new circle
Put x= -λ cosα /2 and y =λ sin α /2 and solve for λ
(-λ cosα /2 ) *cos α + (-λ sin α /2 )* sin α =p
-λ /2 (cos² α +sin² α )=p
-λ/2 =p
λ =-2p
Use this value of λ in eq(3) which will give us the final equation of the new circle.

x² + y² --a² -2p(x cos α +y sinβ -p) =0
Hence proved.