The line x−2y−1=0 intersects the circle x2+y2+4x−2y−5=0 at the points P and Q, then √5PQ is
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Solution
Given circle x2+y2+4x−2y−5=0 ⇒(x+2)2+(y−1)2=10 Now r=√10 and center C(−2,1) length of perpendicular from C on line x−2y−1=0 will be d=|−2−2−1|√5=√5 Hence PQ= length of the chord ∴PQ=2√10−5=2√5 units