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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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@Raskolnikov which 2 solutions? –  vikiiii Apr 4 '12 at 16:21
    
Nope, I made a mistake. The problem leads to a quadratic equation, but one of the solutions implies a rectangle with zero area, which should be discarded. So there's only one left. –  Raskolnikov Apr 4 '12 at 16:23
    
Thanks to @MarkBennet , I deleted my wrong answer. –  user2468 Apr 4 '12 at 18:10

4 Answers 4

up vote 3 down vote accepted

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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can you please explain your equation? –  vikiiii Apr 4 '12 at 16:49
    
@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 $x^2+y^2=r^2$. –  David Mitra Apr 4 '12 at 16:55

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 $(1)$ $$r^2=(40)^2+(30)^2$$ $$r=50\rm{cm}.$$

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