Bunuel wrote:

The circle with center C shown above is tangent to both axes. If the distance from O to C is equal to k, what is the radius of the circle, in terms of k ?(A) k

(B) \(\frac{k}{\sqrt{2}}\)

(C) \(\frac{k}{\sqrt{3}}\)

(D) \(\frac{k}{2}\)

(E) \(\frac{k}{3}\)

Look at the diagram below:

Attachment:

Circle2.png

Since OC=k, then \(r^2+r^2=k^2\) --> \(r=\frac{k}{\sqrt{2}}\).

Answer: B.

Bunuel shouldnt we rationalise denominator

\(r=\frac{k}{\sqrt{2}}\).

for instance if we have \(\frac{\sqrt{6}}{\sqrt{2}}\) = \(\frac{\sqrt{6}}{\sqrt{2}}\) *\(\frac{\sqrt{2}}{\sqrt{2}}\) = \(\frac{\sqrt{12}}{\sqrt{4}}\)

so we get \(\frac{\sqrt{12}}2\)

As you see in the denominator 2 is without radical sign, so why you didn't write it so ?

Another question sqrt of 2 is 1.4 you could write it so in the denominator too ? in which case you could write 1.4 ? can you please explain ?

pleaseee:)