# 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

1
by 2092000

Log in to add a comment

by 2092000

Log in to add a comment

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

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