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CBSE Class 12-science Answered

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Asked by chakrisampathhanuman | 25 Apr, 2020, 08:44: PM
answered-by-expert Expert Answer
Q: The locus of the foot of the perpendicular drawn from origin to a variable line passing through fixed point (2, 3) is a circle whose diameter is
Solution:
Let P(x, y) be any point on the Locus.
The given line passes through Q(h, k).
Since P and Q lie on the same line, they are collinear and Q is the foot of the perpendicular from origin O.
i.e. PQ is perpendicular to OP.
Slope of PQ = y-3/x-2
Slope of OP = y/x
Product of slope of perpendicular lines = -1
(y-3/x-2)(y/x)=-1
x2+y2-2x-3y=0
x2+y2-2x-3y+1+9/4-1-9/4=0
(x-1)2+(y-3/2)2=13/4
R a d i u s space o f space t h e space c i r c l e equals square root of 13 over 4 end root equals fraction numerator square root of 13 over denominator 2 end fraction
D i a m e t e r space o f space t h e space c i r c l e equals square root of 13
Answered by Renu Varma | 26 Apr, 2020, 09:37: PM
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