JEE Advance - Mathematics (2009 - Paper 1 Offline - No. 13)
Tangents drawn from the point P (1, 8) to the circle
$${x^2}\, + \,{y^2}\, - \,6x\, - 4y\, - 11 = 0$$
touch the circle at the points A and B. The equation of the cirumcircle of the triangle PAB is
$${x^2}\, + \,{y^2}\, - \,6x\, - 4y\, - 11 = 0$$
touch the circle at the points A and B. The equation of the cirumcircle of the triangle PAB is
$${x^2}\, + \,{y^2}\, + \,4x\,\, - 6y\, + 19 = 0$$
$${x^2}\, + \,{y^2}\, - \,4x\,\, - 10y\, + 19 = 0$$
$${x^2}\, + \,{y^2}\, - \,2x\,\, + 6y\, - 29 = 0$$
$${x^2}\, + \,{y^2}\, - \,6x\,\, - 4y\, + 19 = 0$$
توضیح
From the given data, the centre of the circle is C(3, 2).
Since, CA and CB are perpendicular to PA and PB, CP is the diameter of the circumcircle of triangle PAB. Its equation is
$$(x - 3)(x - 1) + (y - 2)(y - 8) = 0$$
or $${x^2} + {y^2} - 4x - 10y + 19 = 0$$.