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Question

Tangents are from the point (h,k) to the circle x2+y2=a2;then the the area of the triangle formedby the straight line joining their points of contact is a(h2+k2−a2)32h2+k2.

A
True
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B
False
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Solution

The correct option is A True

Let tangents are drawn from P(h,k) to circle x2+y2=a2 at point of contact of tangents be Q and R

Equation of chord of contact QR is

hx+ky=a2

hx+kya2=0 (1)

Draw PS perpendicular to QR

PS=h2+k2a2h2+k2
length of tangent =PR= S1

⇒PR= h2+k2a2

In △PSR , PS2+SR2=PR2

SR2=(h2+k2a2)2(h2+k2a2h2+k2)2

SR2=(h2+k2a2)(1h2+k2a2h2+k2

SR2=(h2+k2a2)(h2+k2h2k2+a2h2+k2

SR2=(h2+k2a2)(a2h2+k2

SR2=a(h2+k2a2)h2+k2

But QR=2SR

QR=2ah2+k2a2h2+k2

Area of △PQR=12(QR)(PS)

=12×2ah2+k2a2h2+k2×h2+k2a2h2+k2

Area=a(h2+k2a2)32h2+k2

1916674_1031348_ans_e53ddd81d7624715bced510e48350df6.png

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