Introduction to...

$${ \left( x-2 \right)  }^{ 2 }+{ \left( y+3 \right)  }^{ 2 }=16$$ touching the ellipse $$\dfrac { { \left( x-2 \right)  }^{ 2 } }{ { p }^{ 2 } } +\dfrac { { \left( y+3 \right)  }^{ 2 } }{ { q }^{ 2 } } =1$$ from inside. If $$\left( 2,-6 \right)$$ is one focus of the ellipse then $$\left( p,q \right)=$$

If tangent at (1, 2) to the circle $$c_1$$ : $$x^2$$ + $$y^2$$ = 5 intersects the circle $$c_2$$ : $$x^2$$ + $$y^2$$ = 9 at A & B and tangents at A & B to the second circle meet at point C, then the co-ordinates of C is

The vertices of a triangle are $$\left( pq,\cfrac { 1 }{ pq }  \right) ,\left( qr,\cfrac { 1 }{ qr }  \right) ,\left( rp,\cfrac { 1 }{ rp }  \right) $$ where $$p,q,r$$ are roots of the equation $${ y }^{ 3 }-3{ y }^{ 2 }+6y+1=0$$. The coordinates of its centroid are

Consider three lines y axis, y = $$2$$ and lx + my = 1 where (l, m) lies on $$y^2 = 4x$$. Locus of circum centre of triangle formed by given three lines is a parabola whose vertex is

If image of origin along plane mirror passing through $$P,A,B$$ is $$\left( \alpha ,\beta ,\gamma  \right) $$ then

One of the end-points of a circle having centre at origin is $$A(3,-2)$$, then the other end-point of the diameter has the coordinates

If $$\left( -2,-1 \right) ,\left( 1,0 \right) ,\left( x,3 \right) ,\left( l,y \right) $$ form a parallelogram the $$\left( x,y \right) =$$\left[  \right] $$

The intercept cut off by a line from y-axis is twice that of the intercept cut off from the x-axis. If line passes through (1,2), then the equation of the line is 

If the line segment joining the point $$P(x_1, y_1)$$ and $$Q (x_2, y_2)$$ subtends an angle $$\alpha$$ at the origin $$O$$. Then which of the following is true,

If $$\left( { x }_{ 1 },{ y }_{ 1 } \right) ,\left( { x }_{ 2 },{ y }_{ 2 } \right) ,\left( { x }_{ 3 },{ y }_{ 3 } \right) $$ are the vertices of an equilateral triangle such that $$\\ \left( { x }_{ 1 }-2 \right) ^{ 2 }+\left( { y }_{ 1 }-3 \right) ^{ 2 }=\left( { x }_{ 2 }-2 \right) ^{ 2 }+\left( { y }_{ 2 }-3 \right) ^{ 2 }=\left( { x }_{ 3 }-2 \right) ^{ 2 }+\left( y_{ 3 }-3 \right) ^{ 2 }$$ then $${ x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }+2\left( { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 } \right) =\\ $$