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The rectangle at the corner measures 10 cm * 20 cm.

The right bottom corner of the rectangle is also a point on the circumference of the circle.

What is the radius of the circle in cm? Is the data sufficent to get the radius of circle?

enter image description here

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    Tha$n$ks to @MarkBennet , $I$ deleted my wrong answer.2012-04-04

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Hint: with a coordinate system at the center of the circle, the point of intersection of the circle with the rectangle is $(10-r,r-20)$, so $ (10-r)^2+(r-20)^2=r^2. $

Note also, that to be in the situation imposed by the diagram, you must have $r>20$.

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    @vikiiii The equation of a circle with center $(a,b)$ ($(0,0)$ here) and radius $r$ is $(x-a)^2+(y-b)^2=r^2$. We know the point $(x,y)=(10-r, r-20)$ is on our circle; so, this point satisfies the equation $x^2+y^2=r^2$.2012-04-04
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Let $R$ denote the radius, and let $w$ and $h$ be the width and the height of the rectangle.

Consider right triangle, formed by the center of the circle $O$, point where the rectangle touches the circle $A$ and the point $B$ - projection of $A$ on the horizontal diameter.

Then, by Pythagorean theorem: $ \begin{eqnarray} R^2 &=& (R-w)^2 + (R-h)^2 \\ R^2 &=& 2 R^2 - 2 R(w+h) + w^2 + h^2 \end{eqnarray} $ It remains to solve this quadratic equation, and choose the appropriate root (considering the special case of a square, when $w=h$, helps): $ R = w + h + \sqrt{2 w h} $

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enter image description here

In the circle shown above the triangles $\triangle AGT$ and $\triangle TKX$ are similar.

We know $BC=10$ and $AG=20$

Let $CK=y$ and radius of the circle $BX=R$

In the similar triangles $\triangle AGT$ and $\triangle TKX$ we have,

$\frac{AG}{GT}=\frac{TK}{KX}$

$\frac{20}{10+y}=\frac{R-20}{R-(10+y)}$

i.e. $CK=y=10$

$GT=BK=BC+CK=10+10$

$AT=\sqrt{{AG}^2+{GT}^2}=20\sqrt{2}$

${TK}^2+{KX}^2={TX}^2$

$(R-20)^2+(R-20)^2={TX}^2$

TO BE CONTINUED

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$r^2=x^2+y^2 \tag{1}$ $r=y+20$ and $r=x+10$ therefore $y+20=x+10 \quad \mbox{then } \quad y=x-10 \tag{2}$ Substitute $(2)$ into $(1)$ $(r+10)^2=x^2+(x-10)^2$ $x^2+20x+100=x^2+x^2-20x+100$ $X^2=40x$ $x=40$ Then substitute $x=40$ into $(2)$ $y=40-10$ $y=30$ substitute $x=40$ and $Y=30$ into $(1)$ $r^2=(40)^2+(30)^2$ $r=50\rm{cm}.$