It is simple if you think about it

Geometry Level 3

If a line is drawn through a fixed point P ( α , β ) P(\alpha,\beta) to the circle x 2 + y 2 = a 2 x^2 + y^2 = a^2 at A A and B B , then express P A × P B PA \times PB in terms of α , β \alpha,\beta and/or α \alpha .

α 2 + β 2 − a 2 \alpha^2 + \beta^2 - a^2 a 2 a^2 α 2 + β 2 + a 2 \alpha^2 + \beta^2 + a^2 α 2 + β 2 \alpha^2 + \beta^2

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Vishwak Srinivasan by Vishwak Srinivasan , 24, India

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2 solutions

From the image above and using Tangent-Secant theorem, it is known that

P A × P B = P T 2 PA \times PB = PT^2

Using notation as found in S.L.Loney's Coordinate geometry (to avoid ambiguity):

The length of the tangent, i.e., P T = S 1 PT = \sqrt{S_1}

S 1 = α 2 + β 2 − a 2 \sqrt{S_1} = \sqrt{\alpha^2 + \beta^2 - a^2}

P A × P B = P T 2 = S 1 = α 2 + β 2 − a 2 PA \times PB = PT^2 = S_1 = \boxed{\alpha^2 + \beta^2 - a^2}

Note

  • S 1 S_1 is nothing but a point substituted in the circle S = 0 S = 0 , i.e., x 2 + y 2 − a 2 = 0 x^2 + y^2 - a^2 = 0
4 Helpful 0 Interesting 0 Brilliant 0 Confused

If one was pressed for time, maybe one could assume ( α , β ) (\alpha, \beta) to be a point on the circle. Then, α 2 + β 2 = a 2 ⋅ \displaystyle \alpha^{2} + \beta^{2} = a^{2} \cdot

Moreover, P A × P B = 0 \displaystyle PA \times PB = 0 and the result follows from the options.

Nice Solution by the way ! Cheers

B.S.Bharath Sai Guhan - 5 years, 3 months ago

Draw a diameter A 0 B 0 A_0 B_0 passing through P.

P A ∗ P B = P A 0 ∗ P B 0 PA*PB=PA_0*PB_0

from the diagram we can see that P A 0 = r − a PA_0=r-a & P B 0 = r + a PB_0=r+a where r = α 2 + β 2 r=\sqrt{\alpha^2+\beta^2}

So our equation now becomes

P A ∗ P B = ( r − a ) ( r + a ) = r 2 − a 2 = α 2 + β 2 − a 2 PA*PB=(r-a)(r+a)=r^2-a^2=\alpha^2+\beta^2-a^2

2 Helpful 0 Interesting 0 Brilliant 0 Confused

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