If one end of a diameter of the circle $x^2 + y^2 - 3x + 8y - 4 = 0$ is $(6, -3)$,then what is the other end of the diameter?

Consider a semicircle of radius $1$ unit constructed on the diameter $AB$ and let $O$ be its centre. Let $C$ be a point on $AO$ such that $AC:CO = 2:1$. Draw $CD$ perpendicular to $AO$ with $D$ on the semicircle. Draw $OE$ perpendicular to $AD$ with $E$ on $AD$. Let $OE$ and $CD$ intersect at $H$. Then,$DH$ equals

Let $A(3,5)$,$B(-2,-7)$ and $C(\alpha, \beta)$ be three points such that $\angle ACB$ is a right angle and the area of triangle $ABC$ is $\frac{82}{3}$ square units. Then the number of such points $C$ is

Let $PA$ and $PB$ be the tangent segments drawn from point $P(6, 8)$ to the circle with the centre at origin $O(0, 0)$. The radius $r$ of the circle for which the area of quadrilateral $PAOB$ is maximum, is...

Position of the point $(1, 1)$ with respect to the circle $x^2 + y^2 - x + y - 1 = 0$ is

If a circle passes through the points of intersection of the lines $\lambda x - y + 1 = 0$ and $x - 2y + 3 = 0$ with the coordinate axes,then the value of $\lambda$ is: