Question

given the circles x²+y²-4x-5=0 and x²+y²+6x-2y+6=0 let P be a point (alpha , bita ) such that the tangent from P to both the circles are equal , then​

Answer 1

Answer:

  ...then  10α - 2β + 11 = 0.

Step-by-step explanation:

In general, a circle is given by an equation of the form

   f(x, y)  =  (x - a)² + (y - b)² - r²  =  0

where (a, b) is the centre C and r is the radius.

Consider a point Q with coordinates (x, y) and let a tangent to the circle from Q touch the circle at T.  Then CQT is a right angle triangle with a right angle at T.  By Pythagoras' Theorem,

   QT² = QC² - CT²    ⇒    QT²  =  f(x, y)

So the value of f(x, y) is just the square of the length of a tangent from Q.

In this problem, there are two circles, one with equation

   f(x, y) = x² + y² - 4x - 5  =  0

and one with equation

   g(x, y) = x² + y² + 6x - 2y + 6  =  0.

Since the point P = (α, β) is such that tangents from P to both circles are equal, this means

    f(α, β) = g(α, β)

⇒  α² + β² - 4α - 5  =  α² + β² + 6α - 2β + 6

⇒  10α - 2β + 11  =  0.

Hope this helps!