The locus of the mid points of the chords of the circle x2-y2-a x-b y-0 which subtend a right angle at a-2- b-2 is -A- a x-b y-0B- a x-b y-a2-b 2C- x2-y2-a x-b y-a2-b2-8-0x2-y2-a x-b y-a2-b2-5-0

The correct option is D C= ( a/2,b/2 ),r=1/2√(a^2+b^2) P = (x, y) cos a/4=CP/r ⇒ CP^2 = r^2 cos^2 π/2 ( X a/2 )^2+ ( Y b/2 )^2=a^2+b^2/8 ...

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Standard XII

Mathematics

Equation of a Chord Joining Two Points with Circle in Parametric Form

The locus of ...

Question

The locus of the mid points of the chords of the circle $x^{2} + y^{2} - a x - b y = 0$ which subtend a right angle at $(\frac{a}{2}, \frac{b}{2})$ is :

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Solution

The correct option is C

$C = (\frac{a}{2}, \frac{b}{2}), r = \frac{1}{2} \sqrt{a^{2} + b^{2}}$

P = (x, y)

$c o s \frac{a}{4} = \frac{C P}{r} \Rightarrow C P^{2} = r^{2} c o s^{2} \frac{π}{2} {(X - \frac{a}{2})}^{2} + {(Y - \frac{b}{2})}^{2} = \frac{a^{2} + b^{2}}{8}$

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The locus of the mid points of the chords of the circle x2+y2−ax−by=0 which subtend a right angle at (a2,b2) is :

The correct option is C C=(a2,b2),r=12√a2+b2 P = (x, y) cosa4=CPr⇒CP2=r2 cos2 π2(X−a2)2+(Y−b2)2=a2+b28

The locus of the mid points of the chords of the circle $x^{2} + y^{2} - a x - b y = 0$ which subtend a right angle at $(\frac{a}{2}, \frac{b}{2})$ is :

The correct option is C

$C = (\frac{a}{2}, \frac{b}{2}), r = \frac{1}{2} \sqrt{a^{2} + b^{2}}$

P = (x, y)

$c o s \frac{a}{4} = \frac{C P}{r} \Rightarrow C P^{2} = r^{2} c o s^{2} \frac{π}{2} {(X - \frac{a}{2})}^{2} + {(Y - \frac{b}{2})}^{2} = \frac{a^{2} + b^{2}}{8}$