Finding the radius of the circle confusion

Question

Consider the figure below with $B$ as the point of tangency and $AB=BC=4$. Find radius.

My try:

What I did is, at $B$ draw a perpendicular which obviously passes through center of the circle $O$. Now after that I am getting a contradictory radii as shown below.

Since $\Delta OBC$ is $45-45-90$ triangle with $BC=4$, we have radius as $OB=4$. But $\Delta OMD$ is $45-45-90$ too with $OM=MD=4$, so the radius is $4\sqrt{2}$.

Am I missing anything?

Please pardon me for figure being not to scale.

Answer 1

First complete your triangle: the top point is called $D$.

You can find the centre of a circle on the perpendicular bisector of two points, so you define the perpendicular bisector of $BD$. When you put $A$ on coordinate $(0,0)$, $B$ on $(4,0)$ and so on, you find the equation of that perpendicular bisector as:

$y=\frac{x}{2} + 3$

Another line, going through the centre, is the perpendicular line going through $B$, which has the following equation:

$x=4$

Both lines intersect at $(4,5)$, which is the centre of your circle.
The distance between $(4,5)$ and $B$ equals $5$, which is the radius, hence:

$$r=5$$

In a drawing:

It can also be calculated in a purely geometrical way:

Let's consider the two triangles $ABD$ and $FGB$:

• Both are right-angled triangles.
• Angles $ADB$ and $FBG$ are equal (two angles, cut of between parallel lines ($AD$ and $BG$), are equal).

=> both triangles are similar.

Due to their similarity, the ratios of their sides are equal:

$\frac{|BD|}{|AD|} = \frac{|GB|}{|FB|}$

1. $|AD| = 8$
2. $|BD| = \sqrt{|AD|^2 + |AB|^2} = \sqrt{80}$
3. $|FB| = \frac{|BD|}{2} = \frac{\sqrt{80}}{2}$

Hence:

$$|GB| = \frac{\sqrt{80} \cdot \sqrt{80}}{8 \cdot 2} = \frac{80}{16} = 5$$

Answer 2

Let $AD$ intersect the circle at the point $E$

$AD=AC=8$

$AB^2=AE \times AD$ (Power of a point)

$4^2=AE\times 8$

$AE=2$

Draw $OM\perp AD$ and $OB\perp AB=r$

$DM=EM=3$ ($\perp $line from centre bisects chord) Since $ABOM$ is a rectangle, $OB=r=AM=5$