Question

Tangents PT1 and PT2 are drawn from point P to the circle x^2+y^2=a^2. If the point P lies on the line px+qy+r=0, then locus of circumcircle PT1T2 is

Answer 1

Answer:

px + qy + r/2 = 0

Step-by-step explanation:

Given that Tangent PT1 and PT2 are drawn from point P to the circle

x² + y² = a²,

Now,

let the point P be (h, k).

So, T1T2 will become the polar of the circle,hence Equation of T1T2 is given by S1 = 0, which is

hx + ky -a² = 0.

Hence , equation of line T1T2 is hx + ky - a² = 0

We need to find equation of circumcircle of PT1T2,

but we now that equation of any circle which passes through points of

intersection of a given circle (S) and a given line (L) is given by

S+λL = 0

So, equation of any circle passing through T1T2 will be in the form

(x² + y² - a²) + λ(hx + ky - a²) = 0

But this circle should pass through point P(h, k)

=> (h² + k² -a²) + λ(h² + k² - a²) = 0

=> λ = -1

Thus, equation of circumcircle of triangle PT1T2 is given by

x² + y² - a² - (hx + ky - a²) = 0

=> x² + y² -hx - ky = 0,

Center of the circum circle, let it be (α, β) = (h/2, k/2)

=> h = 2α, k = 2β.

But given that point P(h, k) lies on the line px + qy + r = 0

=> p(2α) + q(2β) + r = 0

=>pα + qβ + r/2 = 0

For any general point (x, y):

px + qy + r/2 = 0