# Q.No-39)  From a point P(1,0) , a tangent PA is drawn to the circle x2+y2-8x+12=0 , A being the point of contact. Find  the equations of the tangents to the circle from the middle point of PA if A is in the first quadrant.41) A tangent is drawn to the circle (x-a)2 + y2 = b2 and a perpendicular tangent to the circle (x+a)2 + y2 = c2 . Find the locus of their intersection and prove that the bisectors of the angles between them always touch one or other of two other fixed circles.90) The base of a triangle through a fixed point (a,b) and its sides are respectively bisected at right angles by the lines y2-8xy-9x2=0. Prove that the locus of the vertex is a circle. Find its Equation.

357 Points
14 years ago

39. Dear Student,

Please have a look at the symmetry of the problem.

First let us find the coordinates of A(x, y).

One equation is –

(4–1)2 + (0–1)2 = (x–1)2 + (y–0)2 + (4–x)2 + (0–y)2                 …. (i)

Second equation is –

Slope of PA × slope of AL + –1                                              ….. (ii)

Solving these two equations you get the coordinates of A(x, y)

Then by applying mid-point formula, we get the coordinates of mid-point of PA say (x1, y1)

Then we get the equation of pair of tangents to the circle from (x1, y1) as

?S1 = T2

Where       S = x2 + y2 + 2fx + 2fy + c

S1 = S12 + y12 + 2fx1 + 2fy1 + c

and   T = x x1 + y y1 + y(x + x1) + f(y + y1) + c