The centre of a circle is `C(2,k)`.If `A(2,1)` and `B(5,2)` are two points on its circumference , then the value of k is

Let A(-3, 2) , B(4,1) and C(-2,k) be three points such that AC = BC . Find the value of k .

Consider a family of circles which are passing through the point (-1,1) and are tangent to the x-axis.If (h,k) are the coordinates of the center of the circles,then the set of values of k is given by the interval.k>=(1)/(2)(b)-(1)/(2)<=k<=(1)/(2)k<=(1)/(2)(d)0

The value of k for which the points A(0, 1), B(2, k) and C(4, -5) are colinear is :

The value of k for which the points A(0, 1), B(2, k) and C(4,-5) are colinear is:

Show that the points (-3, 2), (2, -3) and (1, 2 sqrt(3) ) lie on the circumference of that circle, whose centre is origin.

Show that the points (-3, 2), (2, -3) and (1, 2 sqrt(3) ) lie on the circumference of that circle, whose centre is origin.

Find the value of k, if the points A(7,-2),B(5,1) and C(3,2k) are collinear.

If the point C(k, 4) divides the join of A(2, 6) and B(5, 1) in the ratio 2:3 then find the value of k.

If the point C(k, 4) divides the join of A(2, 6) and B(5, 1) in the ratio 2:3 then find the value of k.