A rod PQ of length 2 a slides with its ends on the axes the locus of the circum-centre of - OPQ isA- x2- y2-4 a2B- x2- y2-2 a2C- x2- y2-3 a2D- x2-y2-a2

The correct option is D + = Let C (h, k) be a point in the locus C(h, k) is mid point of PQ ⇒ P(2h, 0), Q(0, 2k) PQ = 2a ⇒ PQ^2 = 4a^2 ⇒ 4h^2 + 4k^2 = 4a^2 ⇒ ...

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Standard XII

Mathematics

Intercept Made by Circle on Axes

A rod PQ of l...

Question

A rod PQ of length 2a slides with its ends on the axes the locus of the circum-centre of $Δ$ OPQ is

= 2

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= 4

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= 3

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Solution

The correct option is D + =
Let C (h, k) be a point in the locus
C(h, k) is mid point of PQ $\Rightarrow$ P(2h, 0), Q(0, 2k)
PQ = 2a $\Rightarrow P Q^{2} = 4 a^{2}$ $\Rightarrow 4 h^{2} + 4 k^{2} = 4 a^{2}$ $\Rightarrow h^{2} + k^{2} = a^{2}$
Locus of (h, k) is $x^{2}$ + $y^{2}$ = $a^{2}$

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A rod PQ of length 2a slides with its ends on the axes the locus of the circum-centre of Δ OPQ is

The correct option is D + = Let C (h, k) be a point in the locus C(h, k) is mid point of PQ ⇒ P(2h, 0), Q(0, 2k) PQ = 2a ⇒PQ2=4a2 ⇒4h2+4k2=4a2 ⇒h2+k2=a2 Locus of (h, k) is x2 + y2 = a2

A rod PQ of length 2a slides with its ends on the axes the locus of the circum-centre of $Δ$ OPQ is

The correct option is D + =
Let C (h, k) be a point in the locus
C(h, k) is mid point of PQ $\Rightarrow$ P(2h, 0), Q(0, 2k)
PQ = 2a $\Rightarrow P Q^{2} = 4 a^{2}$ $\Rightarrow 4 h^{2} + 4 k^{2} = 4 a^{2}$ $\Rightarrow h^{2} + k^{2} = a^{2}$
Locus of (h, k) is $x^{2}$ + $y^{2}$ = $a^{2}$