Question

PQ and RS are two chords of a circle with centre O. The perpendiculars drawn from O to PQ and RS are OX and OY respectively. Show that PQ² - RS² = 4OY - 4OX.​ ​

Answer 1

Answer:

• See below

Step-by-step explanation:

According to the picture we have:

• O - center of the circle
• OQ and OS are same as radius of circle
• OX is the perpendicular bisector of PQ
• OY is the perpendicular bisector of RS

Consider right triangles OXQ and OYS.

Their hypotenuse are the radius:

• OQ = OS (radius)

Use Pythagorean theorem:

• OQ² = OX² + XQ² ⇒ OQ² = OX² + (PQ/2)² ⇒ OQ² = OX² + PQ²/4
• OS² = OY² + YS² ⇒ OS² = OY² + (RS/2)² ⇒ OS² = OY² + RS²/4

Since OQ = OS, their squares are also equal:

• OX² + PQ²/4 = OY² + RS²/4
• 4OX² + PQ² = 4OY² + RS²
• PQ² - RS² = 4OY² - 4OX²
• Proved