Question

If the length of tangents from
(a,b) to the circles x' + y² - 4x – 5 = 0
and x² + y² +6x – 2y+6=0 are equal,
then​

Answer 1

Answer:

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Answer 2

Correct question is "If the length of tangents from (a,b) to the circles x' + y² - 4x – 5 = 0 and x² + y² +6x – 2y+6=0 are equal then"

Answer:

If the length of tangents from (a,b) to the circles x' + y² - 4x – 5 = 0 and x² + y² +6x – 2y+6=0 are equal then 10a-2b=-11.

Step-by-step explanation:

Given circles are x² + y²-4x-5= 0.....(1) and x² + y²+6x-2y+6=0.....(2) Point from where length of tangents are equal lies on the radical axis.

Now question is what is radical axis ?

Equation of the radical Axis

=> x(g-g')) + y(f – f') + c – c' = 0. b. If S' = 0 and S = 0 touch each other, then the equation of the common tangent to the two circles at by the point of contact is given by S-S'=0.

We are comparing equation (1) with x²+y²+2gx+2fy+c=0

and equation (2) with x²+y²+2g'x+2f'y+c=0 and get

g =(-4) and g' = 6

f = 0 and f' = -2

So the radical axis is

S-S'= 0

$x( - 4 - 6) + y(0 + 2) + ( - 5 - 6) = 0 \\ x( - 10) + y(2) + ( - 11) = 0 \\ - 10x + 2y - 11 = 0 \\ 10x - 2y+ 11 = 0$

Putting (a, b) in the equation of radical axis and we get,

$10a - 2b + 11 = 0 \\ 10a - 2b = - 11$

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