two circles of radii x cm and y cm(x > y) intersect at two points P and Q respectively.if the distance 'd' between the centres of two circles is given by D^2=x^2-y^2,prove that length of the common chord is 2y cm
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See diagram.
Given C1 P = x, C2 P = y, PR = common chord.
C1 C2 = d . Let the point of intersection of PR with C1 C2 be Q.
Given C1C2² = d² = x² - y²
In the Δ C1C2P, we have
C1P² = C2P² + C1C2² because, x² = y² + (x² - y²)
Hence ΔC1C2P is a right angle triangle at C2. Q and C2 are the same point. Hence ∠C1C2R and ∠C1C2P are right angles. Hence, PR is same as the straight line PC2R.
So the length of PC2R = PC2+ C2R = y + y = 2y.
Given C1 P = x, C2 P = y, PR = common chord.
C1 C2 = d . Let the point of intersection of PR with C1 C2 be Q.
Given C1C2² = d² = x² - y²
In the Δ C1C2P, we have
C1P² = C2P² + C1C2² because, x² = y² + (x² - y²)
Hence ΔC1C2P is a right angle triangle at C2. Q and C2 are the same point. Hence ∠C1C2R and ∠C1C2P are right angles. Hence, PR is same as the straight line PC2R.
So the length of PC2R = PC2+ C2R = y + y = 2y.
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