Show that inverse of a point a with
respect to the circle `|z-c|=R(aandc` are complex
numbers, centre and redius R) is the point `c+(R^(2))/(bara-barc),`

The inverse point of P(4,3) W.r.t the circle x^(2)+y^(2)=25 is

The figure shows a fixed circle C_1 with equation (x-1)^2 +y^2 = 1 and a shrinking circle C_2 with radius r and centre at the origin. P is the point (0,r), Q is the upper point of intersection of the two circles and R is the point of intersection of the line PQ and the x-axis. The coordinates of the point R as C, shrinks ie. as r->0 is

The area of the region represented by r<=|z-2|<=5 where z is a complex number is 9 pi then the value of r is

If |(z-z_1)/(|z-z_2|)=3,w h e r ez_1a n dz_2 are fixed complex numbers and z is a variable complex number, then z lies on a (a).circle with z_1 as its interior point (b).circle with z_2 as its interior point (c).circle with z_1 as its exterior point (d).circle with z_2 as its exterior point

P and Q are the middle points of two chords (not diameters) AB and AC respectively of a circle with centre at a point O. The lines OP and OQ are produced to meet the circle respectively at the points R and S. T is any point on the major arc between the points R and S of the circle. If angle BAC = 32^(@), angle RTS = ?

Three points represented by the complex numbers a,b,c lie on a circle with centre 0 and rdius r. The tangent at C cuts the chord joining the points a,b and z. Show that z= (a^-1+b^-1-2c^-1)/(a^-1b^1-c^2)

IF OABC is a tetrahedron where O is the origin and A,B and C have respective positions vectors as bara,barb and barc then prove that the circumcentre of the tetrahedron is ((bara)^2(barb times barc)+(barb)^2 (barc times bara)+(barc)^2 (bara times barb))/(2[bara barb barc])

From a point P, two tangents PA and PB are drawn to a circle with centre O and radius r. If OP=2r, show that triangleAPB is equilateral.