Two tangents are drawn from the point P(-1, 1) to the circle $x^2 + y^2 - 2x - 6y + 6 = 0$. If these tangents touch the circle at points A and B, and if D is a point on the circle such that length of the segments AB and AD are equal, then the area of the triangle ABD is equal to :
Show Hint
The chord of contact from an external point to a circle is obtained using the $T=0$ form.
Once the contact points are known, symmetry and distance constraints often reduce the
problem to simple coordinate geometry.
The given circle is: \[ x^2 + y^2 - 2x - 6y + 6 = 0 \] Comparing with the standard form $x^2+y^2+2gx+2fy+c=0$, we get: \[ g=-1,\quad f=-3,\quad c=6 \] Hence, Center $C = (1,3)$ Radius $r = \sqrt{g^2+f^2-c} = \sqrt{1+9-6} = 2$ Step 1: Equation of chord of contact AB The point from which tangents are drawn is $P(-1,1)$. The equation of the chord of contact from $(x_1,y_1)$ to the circle is: \[ xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0 \] Substitute $x_1=-1$, $y_1=1$, $g=-1$, $f=-3$, $c=6$: \[ -x + y - (x-1) - 3(y+1) + 6 = 0 \] Simplifying: \[ -2x - 2y + 4 = 0 \quad \Rightarrow \quad x + y - 2 = 0 \] This is the equation of chord AB. Step 2: Coordinates of points A and B Solve the system: \[ x + y = 2 \] \[ x^2 + y^2 - 2x - 6y + 6 = 0 \] Substitute $y=2-x$ into the circle equation: \[ x^2 + (2-x)^2 - 2x - 6(2-x) + 6 = 0 \] \[ 2x^2 - 2 = 0 \Rightarrow x^2 = 1 \] Thus, \[ A(1,1),\quad B(-1,3) \] Step 3: Length of chord AB \[ AB = \sqrt{(1+1)^2 + (1-3)^2} = \sqrt{4+4} = 2\sqrt{2} \] Step 4: Coordinates of point D Let $D(x,y)$ be a point on the circle such that $AD = AB = 2\sqrt{2}$. Equations: \[ (x-1)^2 + (y-3)^2 = 4 \quad \text{(on the circle)} \] \[ (x-1)^2 + (y-1)^2 = 8 \quad \text{(distance } AD) \] Subtracting: \[ (y-1)^2 - (y-3)^2 = 4 \] \[ 4y - 8 = 4 \Rightarrow y=3 \] Substitute in circle equation: \[ (x-1)^2 = 4 \Rightarrow x=3 \text{ or } -1 \] $x=-1$ gives point $B$, hence \[ D = (3,3) \] Step 5: Area of triangle ABD Vertices: \[ A(1,1),\ B(-1,3),\ D(3,3) \] Using determinant formula:
