Let x−2y+7=0 be a chord of the circle x2+y2−2x−10y+1=0. If the midpoint of the chord is P(α,β), then the value of 5|α−β| is
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Solution
Given chord is x−2y+7=0 P=(α,β)
Now, α−2β+7=0⋯(1) CP is perpendicular to chord, so β−5α−1=−2
Using equation (1), we get β−5=−2(2β−8)⇒β=215⇒α=75∴5|α−β|=14
Alternate method :
Given circle is x2+y2−2x−10y+1=0
Centre and radius is C=(1,5),r=5 P is foot of perpendicular drawn from (1,5) to x−2y+7=0, so α−11=β−5−2=−(1−10+712+22)⇒α−11=β−5−2=25⇒α=75,β=215∴5|α−β|=14