Question

A circle touches the y axis at the point $ \left( {0,4} \right) $ and cuts the x-axis in a chord of length $ 6 $ units. The radius of the circle is
A. $ 3 $
B. $ 4 $
C. $ 5 $
D. $ 6 $

Answer 1

Hint: In order to find the radius of the circle, initiate with making a rough diagram with some construction if needed, then a triangle is formed. Use of theorems of circle for the line that is passing through the centre and falling on the chord. Finally use the Pythagoras theorem for the triangle if it’s a right-angled triangle, then solve it and get the result.

Complete step-by-step answer:
We are given that the circle touches the y-axis at a point $ \left( {0,4} \right) $ and cuts the x-axis in a chord which is of length $ 6 $ units.
Considering the information above, the diagram we can represent as:

Doing some construction by joining some lines as:

In this diagram, D is the point where the circle touches the x-axis and from $ \left( {2,0} \right) $ and $ \left( {8,0} \right) $ the circle cuts the x-axis into length of $ 6 $ units.
Drawn a line AC whose length is equal to of the line OB they are parallel to each other and a kind of square is formed by the diagram.
So, the length of OB is $ 4 $ units as from Origin the point is $ 4 $ units above.
So, OB is $ 4 $ units.
Since, $ OB = AC $ , so $ OB = AC = 4{\text{ units}} $ .
From D to C, we can see that the distance between them is $ 3 $ units.
So, from this $ CD = 3{\text{ units}} $ .
Now, a triangle is formed that is $ \Delta ACD $ .
From the theorems of Circle, we know that the line from the centre cuts the chord perpendicularly, so $ \angle ACD = {90^ \circ } $ . Therefore, we can apply Pythagoras theorem in it:
From Pythagoras Theorem, in $ \Delta ACD $ we can obtain the equation as:
 $ {\left( {AC} \right)^2} + {\left( {CD} \right)^2} = {\left( {AD} \right)^2} $
Substituting their values known above from the figure:
 $ {\left( 4 \right)^2} + {\left( 3 \right)^2} = {\left( {AD} \right)^2} $
 $ \Rightarrow {\left( {AD} \right)^2} = {\left( 4 \right)^2} + {\left( 3 \right)^2} $
Squaring both the sides:
 $
   \Rightarrow {\left( {AD} \right)^2} = 16 + 9 \\
   \Rightarrow {\left( {AD} \right)^2} = 25 \;
  $
Taking Square Root both the sides, we get:
\[
   \Rightarrow \sqrt {{{\left( {AD} \right)}^2}} = \sqrt {25} \\
   \Rightarrow AD = 5{\text{ units}} \;
 \]
AD is nothing but the radius of the circle.
Therefore, the Radius of the Circle is \[5{\text{ units}}\].
Hence, Option C is correct.
So, the correct answer is “Option C”.

Note: Remember, Pythagoras Theorem is always applied on a right-angled triangle, no other triangle.
Pythagoras theorem states that the sum of the squares of the perpendicular and the base is equal to the square of the hypotenuse.
We can make some constructions in this kind of questions, or diagrams if needed.