Question

Circle question?

Answer 1

`OB` is radius and `ABQ` is tangent .

Hence `angleOBQ=90^@`

Here `OB=OC=OD="radius of the circle"`

`Delta ODC` is isosceles.

So `angleOCD=angleODC=x`

Similarly

`Delta OBC` is isosceles.

So `angleOCB=angleOBC=angleOBQ-2x=90-2x`

So `angle BCD=angle OCB+angle OCD=90-x`

So central `angle BOD=2angle BCD=180-2x`

Now
obtuse `angleBOD=180^@-y` as BODA is a cyclic quadrilateral .

Hence `180-y=180-2x`

`=>y=2x`