If pair of tangents are drawn from any point (P) on the circle x^2+ y^2=1 to the hyperbola x^2/2 -y^2/1 =1, such that locus of circumcentre of triangle formed by pair of tangents and chord of contact is.?

Question

Answer

Concepts:

Pair of tangents from a point on a circle

Chord of contact of a hyperbola

Circumcentre of a triangle

Locus of a point

Solution:

Let P be a point on the circle x^2 y^2=1.

Let the pair of tangents from P to the hyperbola x^2/2 -y^2/1 =1 intersect the hyperbola at points A and B.

Let PQ be the chord of the hyperbola passing through A and B.

Let O be the circumcentre of triangle PAB.

As the chord of contact of the hyperbola is the perpendicular bisector of the line segment joining the points of contact of the tangents from an external point, PQ is the perpendicular bisector of AB.

Therefore, the point O lies on the line PQ.

Let the coordinates of P be (a,b) such that a^2 b^2=1.

Let the equation of the tangent at A be Tx + Ty = 2.

As A lies on the hyperbola, we have (2a^2 - b^2)x - 2ay = 2.

Similarly, the equation of the tangent at B is (2a^2 - b^2)x + 2ay = 2.

Let the coordinates of the point of intersection of the tangents be (h,k).

Substituting (h,k) in the equations of the tangents, we get h^2 - k^2 = 1.

Therefore, the locus of (h,k) is the hyperbola x^2 - y^2 = 1.

Let the point of intersection of PQ and the y-axis be R.

As PQ is the perpendicular bisector of AB, we have AR = RB.

Therefore, the coordinates of R are (0, 1/√2).

Let the equation of PQ be y = mx + c.

As R lies on PQ, we have 1/√2 = m*0 + c.

Therefore, c = 1/√2.

As O lies on PQ, we have k = mh + c.

Substituting h^2 - k^2 = 1, we get h^2 - (m^2 h^2 + 2m/√2 h + 1/2) = 0.

Solving for h, we get h = ± √(1 + m^2)/√2.

Therefore, the locus of O is the circle with centre (0,1/√2) and radius √(1 + m^2)/√2.

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