a- the common tangent to the circle x2+y2=6xand x2+y2+2x=0 forms an........... triangle

b-if 2 distinct chords of circle x2+y2=ax+by r drawn from pt a,b are divided by the x axis in 2;1 find relation b/w a and b

c- find eqn of locus of feet of perpendicular drawn from origin upon a variable chord of a circle which subtend right angle at origin

1)

x²+ y² = 6x  (1) and x² + y²+ 2x = 0  (2)
So the centre of the circle (1) is (x-3)² +(y-0)²  = 9 is (3 ,0) and radius = 3
And centre of the circle (2) is (x +1)² + (y-0)² = 1  is ( -1 ,0) and radius = 1

C₁B + BC₂  = 4 ,
1 + 3 = 4
Hence the circle touches each other externally.
We can find the coordinate of A by using the external division
So A (x₁ ,y₁)  can be found as :
$x1 =\frac{3(-1) - 1\left(3\right)}{3-1} = -3 and y_{1 } =\frac{3\left(0\right)-1\left(0\right)}{3-1} = 0$

And B(x₂ ,y₂)  by internal division .
$x_2 =\frac{3(-1) +1\left(3\right)}{3+1} =0 and y_2 = 0$
So AB = 3
And AP = AQ  =  $\frac{A{T}_1 +A{T}_2}{2} = \frac{\sqrt{2^2-1^2}+\sqrt{6^2-3^2}}{2} =2\sqrt{3}$
And PB  = $\sqrt{A{P}^2-A{B}^2} =\sqrt{12-9} =\sqrt{3}$
So PQ = 2(3^1/2 )

So all three sides are equal , hence the triangle formed is equilateral.