Q.275. Orthogonality of circles related to ellipse.

Question 275.
The tangent at a point P on the ellipse x^2/a^2 + y^2/b^2 = 1 intersects the major axis in T and N is the foot of the perpendicular from P to the same axis .Show that the circle on NT as a Diameter intersects the auxiliary circle orthogonally.

Answer 275.
Let P (acosθ, bsinθ) be any point on the ellipse.
The equation of tangent at P to the ellipase
x^2/a^2 + y^2/b^2 = 1 is
(bcosθ)x + (asinθ) y = ab

Plugging y = 0, x = a/cosθ
=> T = (a/cosθ, 0) and N = (acosθ, 0)

The equation of the circle with NT as diameter is
(x - acosθ) (x - a/cosθ) + y^2 = 0
=> x^2 + y^2 - a(cosθ + 1/cosθ) x + a^2 = 0 ... (1)
The equation of the auxiliary circle is
x^2 + y^2 - a^2 = 0 ... (2)

For the two circles,
x^2 + y^2 + 2gx + 2fy + c = 0 and x^2 + y^2 + 2g'x + 2f'y + c' = 0,
the condition of orthogonality is 2gg' + 2ff' = c + c'.
For the auxiliary circle, g' = 0 and f' = 0
=> 2gg' + 2ff' = 0
and c = a^2 and c' = - a^2 => c + c' = 0 for the two circles.
Thus, 2gg' + 2ff' = 0 = c + c'
=> the circle with NT as diameter and the auxiliary circle are orthogonal to each other.

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