the locus of poles of tangents to the circle x^2+y^2=a^2 with respect to the circle x^2+y^2+2ax-a^2=0 is
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Circle1: x² + y² = a²
let A (x1, y1) be a point on circle 1: so x1² + y1² = a² --- (1)
Tangent T to circle at point A(x1, y1): x * x1 + y * y1 = a² --- (2)
slope of tangent T at A: - x1/ y1
Slope of Normal N to tangent T at A: y1/x1
Circle 2: (x +a)² + y² = 2 a²
radius R = √2 a center O = (-a , 0)
Equation of straight line OQ parallel to Normal N and passing through O (-a, 0) is:
y = (x+a) * y1/x1 -- (3)
OQ = perpendicular distance from O onto the line T
= | x1 *(-a) + y1 * 0 - a² | / √[x1² + y1² ] --- using (2)
= a |x1 + a| / a
= x1 + a as x1 <= a
Let P (x2, y2) be the pole of line T wrt circle 2.
OP² = (x2+a)² + y2² -- (4)
The pole P of tangent T to circle 1, wrt Circle 2, will lie on the Normal N and at a distance OP from center O :
OP = R² / OQ = 2 a² / (x1 + a) --- (5)
As P(x2, y2) lies on OQ, by (3) we get,
y2 = (x2 + a) * y1 / x1 -- (6)
From (4) and (5) we get:
(x2 + a)² + y2² = 4 a⁴ / (x1 + a)²
=> y2² [ a² / y1² ] = 4 a⁴ / (x1 +a)² using (6) and (1)
=> y2² = 4 a² y1² / (x1 + a)²
let us use parametric representation for circle 1: x1 = a cos Ф and y1 = a sin Ф
y2² = 4 a² a² Sin² Ф / (2 a² Cos²Ф/2)² = a² Sin² Ф / Cos²Ф/2
= 4 a² Sin² Ф/2
So y2 = + or - 2a Sin Ф/2 --- (7)
Cos²Ф/2 = 1 - SIn²Ф/2 = (4a² - y2²) /4a²
tan Ф/2 = + - y2 / √(4a² - y2²)
tan Ф = + 2 y2 √(4 a² - y2² ) / [ 4a² - 2 y2²]
CotФ = (2a² - y2²) / [ y2 √ (4a² - y2²) --- (8)
From (6) we have :
(x2 + a) = y2 * cot Ф = (2 a² - y2² ) / √(4a² - y2²)
Now for getting the locus of the pole P(x2, y2), replace them by (x, y):
(x + a)² = (2 a² - y²)² / (4 a² - y²) --- (9)
let A (x1, y1) be a point on circle 1: so x1² + y1² = a² --- (1)
Tangent T to circle at point A(x1, y1): x * x1 + y * y1 = a² --- (2)
slope of tangent T at A: - x1/ y1
Slope of Normal N to tangent T at A: y1/x1
Circle 2: (x +a)² + y² = 2 a²
radius R = √2 a center O = (-a , 0)
Equation of straight line OQ parallel to Normal N and passing through O (-a, 0) is:
y = (x+a) * y1/x1 -- (3)
OQ = perpendicular distance from O onto the line T
= | x1 *(-a) + y1 * 0 - a² | / √[x1² + y1² ] --- using (2)
= a |x1 + a| / a
= x1 + a as x1 <= a
Let P (x2, y2) be the pole of line T wrt circle 2.
OP² = (x2+a)² + y2² -- (4)
The pole P of tangent T to circle 1, wrt Circle 2, will lie on the Normal N and at a distance OP from center O :
OP = R² / OQ = 2 a² / (x1 + a) --- (5)
As P(x2, y2) lies on OQ, by (3) we get,
y2 = (x2 + a) * y1 / x1 -- (6)
From (4) and (5) we get:
(x2 + a)² + y2² = 4 a⁴ / (x1 + a)²
=> y2² [ a² / y1² ] = 4 a⁴ / (x1 +a)² using (6) and (1)
=> y2² = 4 a² y1² / (x1 + a)²
let us use parametric representation for circle 1: x1 = a cos Ф and y1 = a sin Ф
y2² = 4 a² a² Sin² Ф / (2 a² Cos²Ф/2)² = a² Sin² Ф / Cos²Ф/2
= 4 a² Sin² Ф/2
So y2 = + or - 2a Sin Ф/2 --- (7)
Cos²Ф/2 = 1 - SIn²Ф/2 = (4a² - y2²) /4a²
tan Ф/2 = + - y2 / √(4a² - y2²)
tan Ф = + 2 y2 √(4 a² - y2² ) / [ 4a² - 2 y2²]
CotФ = (2a² - y2²) / [ y2 √ (4a² - y2²) --- (8)
From (6) we have :
(x2 + a) = y2 * cot Ф = (2 a² - y2² ) / √(4a² - y2²)
Now for getting the locus of the pole P(x2, y2), replace them by (x, y):
(x + a)² = (2 a² - y²)² / (4 a² - y²) --- (9)
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