If a tangent to the circle x2 + y2 = 1 intersects the coordinate axes at distinct points P and Q, then the locus of the mid-point of PQ is

Engineering

Mathematics

Tangent and Normal to Circle

Question

If a tangent to the circle x² + y² = 1 intersects the coordinate axes at distinct points P and Q, then the locus of the mid-point of PQ is

x² + y² – 16x²y² = 0

x² + y² – 2x²y² = 0

x² + y² – 4x²y² = 0

x² + y² – 2xy = 0

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Let the mid point be S (h, k)

$∴$ P(2h,0) and Q(0,2k) equation of

$PQ : \frac{x}{2 h} + \frac{y}{2 k} = 1$

$∵$ PQ is tangent to circle at R (say)

$∴ QR = 1 \Rightarrow | \frac{- 1}{\sqrt{{(\frac{1}{2 h})}^{2} + {(\frac{1}{2 k})}^{2}}} | = 1$

$\Rightarrow \frac{1}{4 h^{2}} + \frac{1}{4 k^{2}} = 1$

⇒ x² + y² – 4x²y² = 0