Coordinate Geom...

The centroid and two vertices of a triangles are (4,-8), (-9,7), (1,4) then the area of the triangle is 

The centre of a circle passing through the points (0 , 0), (1 , 0) and touching the circle $$x^2 + y^2$$ = 9 

A circle passes through point $$\left( 3 , \sqrt { \frac { 7 } { 2 } } \right)$$ touches the line pair $$x ^ { 2 } - y ^ { 2 } - 2 x + 1 = 0$$. Center of the circle lies inside the circle $$x ^ { 2 } + y ^ { 2 } - 8 x + 10 y + 15 = 0 .$$ Co-ordinate of centre of circle is

The centers of the passing through (0,0) and (1,0) and touching the circle $${ x }^{ 2 }+{ y }^{ 2 }=9$$is 

In $$\triangle ABC,A(1,2);B(5,5),\angle ACB={ 90 }^{ 0 }$$. If area of $$\triangle ABC$$ is to be 6.5 sq. units, then the possible number of points for C is 

A line forms a triangle of area $$54\sqrt { 3 } $$ sq-units with the coordinate axes. Then the equation of the line if the perpendicular drawn from the origin to the line makes an angle of $${ 60 }^{ \circ  }$$ with the x-axis is 

If the area of the triangle formed by the line -y = 0, x + y = 0 and x - c = 0 is 16 units, c is

Let $$A(h,k),B(1,1)$$ and $$C(2,1)$$ be the vertices of a right- angled triangle with $$AC$$ as its hypotenuse. If the area of the triangle is 1, then the set of values which $$k$$ can take is given by

If the tangent at (3,-4) to the circle $${ x }^{ 2 }+{ y }^{ 2 }-4x+2y-5=0$$ cuts the circle $${ x }^{ 2 }+{ y }^{ 2 }-16x+2y+10=0$$ in A and B then the midpoint of AB is

If the distance of given point $$\left( \alpha ,\beta  \right) $$ from each of two straight lines through the origin is d, then the value of $$\left( \alpha y-\beta x \right) ^{ 2 }$$ is equal to