If two circles are such that the centre of one lies on the circumference of the other, then the ratio of the common chord of two circles to the radius of any of the circles is?

Question

Answer

√3 : 1

Answer

Answer

\`` By data, the radii are congruent.

~ Let the centers of intersecting circles be O and T.
OT = 2r

~ Let the common chord be PQ.

• POTQ will form a rhombus; sides being equal to the radii.

• Diagonals OT and PQ bisect at right angles.

• Diagonals split rhombus into four congruent right triangles; right angled at the intersection of diagonals; side as hypotenuse.

• Each right triangle has sides: ½PQ, ½OT and r

⇒ r² = (½PQ)² + (½OT)² { applying Pythagoras Theorem }

⇒ r² = ¼PQ² + ¼OT²

⇒ r² = ¼(PQ² + OT²)

⇒ 4r² = PQ² + r²

⇒ 3r² = PQ²

⇒ √3r = PQ

⇒ √3/1 = PQ/r

||`` Common chord : radius = √3 : 1 ``||

Answer

Introduction:
When two circles are such that the center of one lies on the circumference of the other, they are called intersecting circles. In this case, we can find the ratio of the common chord of the two circles to the radius of any of the circles.

Explanation:
To understand the ratio of the common chord to the radius, let's consider two intersecting circles with centers at points O and P and radii r and R, respectively.

Step 1: Draw the diagram:
Draw the circles with centers O and P. Mark the point of intersection as A. Draw radii OA and OP.

Step 2: Identify the common chord:
The common chord is the line segment that connects the points of intersection of the circles. Let's call the points of intersection B and C.

Step 3: Find the length of the common chord:
We can use the Pythagorean theorem to find the length of the common chord. In triangle OAB, we have:
OA^2 = OB^2 + AB^2

Since OA and OB are radii of the same circle, they are equal to the radius r. Thus, we have:
r^2 = r^2 + AB^2

Simplifying the equation, we get:
AB^2 = r^2 - r^2 = 0

This implies that AB = 0, which means that the points B and C coincide. Therefore, the common chord is actually a tangent to the circle at point A.

Step 4: Find the ratio of the common chord to the radius:
Since the common chord is a tangent to the circle, it is perpendicular to the radius at the point of tangency. Therefore, we can use the tangent-secant theorem, which states that the square of the length of a tangent segment is equal to the product of the lengths of the secant segment and its external part.

In this case, the tangent segment is AB and the secant segment is OP. Thus, we have:
AB^2 = AP * PB

Since AB is 0, the equation becomes:
0 = AP * PB

Since AP and PB are radii of the circles, they are equal to R and r respectively. Therefore, we have:
0 = R * r

Simplifying the equation, we get:
R * r = 0

This implies that either R = 0 or r = 0. However, both R and r are positive values since they represent the radii of the circles. Therefore, we can conclude that the ratio of the common chord to the radius is 0.

Conclusion:
The ratio of the common chord of two circles to the radius of any of the circles is 0 when the center of one circle lies on the circumference of the other. This means that the common chord is non-existent in this case.

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