Question

A circle is inscribed in a quadrant of circle of radius 8 .What is a measure of radius of inscribed circle.

Answer 1

Given that there is a big circle of radius 8.  In one quadrant a small circle is inscribed.

Let us assume that big circle is centred at the origin and smaller circle is in the I quadratnt.

Then the small circle touches both x and y axes and also touches the big circle at the point (8cos 45, 8 sin 45)

=$(4\sqrt{2} , 4\sqrt{2} )$

Since small circle touches x and y axes, if it has radius r, then centre = (r,r)

Equaion of small circle is $[tex](x-r)^2 +(y-r)^2 = r^2$/tex]

Since $(4\sqrt{2} , 4\sqrt{2} )$ lies on this circle we get

$(4\sqrt{2)-r)^2 +(4\sqrt{2)-r)^2 = r^2$/

Simplify to get a quadratic equation

$r^2-32 \sqrt{2} r+32 =0</p> <p>r =\frac{32 \sqrt{2}}-\sqrt{2048-128} {2} =4\sqrt{2} -4$