Two concentric circles have radii a and b (a>b). Find the length of the chord of the larger circle which touches the smaller circle.
Answer:
2√a2−b2
- Let O be the common center of the two circles and AB be the chord of the larger circle which touches the smaller circle at C.
As the radius of the larger circle is a and that of the smaller circle is b, we have OA=a and OC=b.
The situation given can be represented by the image below.
- AB is the tangent to the circle with radius b.
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
So, OC⊥AB.
AB is the chord of the circle with radius a.
The perpendicular drawn from the center of the circle to a chord bisects the chord.
∴ bisects AB \implies C is the mid-point of AB. - Using Pythagoras' theorem in the right- angled triangle ACO, we have \begin{aligned} & OA^2 = OC^2 + AC^2 && \\ \implies & AC = \sqrt { OA^2 - OC^2 } = \sqrt { a^2 - b^2 } \end{aligned} C is the mid-point of AB. So, AB = 2AC = 2 \sqrt { a^2 - b^2 }.
- Thus, the length of the chord of the larger circle which touches the smaller circle is 2 \sqrt { a^2 - b^2 } .