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Equation of circle is  (x+2a)² + (y-a)² = ka²

This is a circle with origin coord ( -2a, a) and radius a√k

(a) when k =4 the radius is 2a with orgin at (-2a , a).

So when x = 0 , y = a and  this point is directly on the right of the circle origin.

Connecting this point to the origin will be a line parallel to x axis, so the tangent of this point must be y axis. 

(b) given k=4,  the tangent of the circle parallel to x-axis occurs at the point 2a from its origin along y axis. 

So these point occurs at y = a+2a= 3a and y = a-2a = -a and x = -2a.

So the coordinates are ( -2a, 3a) and (-2a, -a)

(c) Given k = 5 

when x = 0             

4a² + (y-a)² = 5a²

y=0,

so origin is a point on the circle.

(d) the line that joint the origin of the circle (-2a,a) and (0,0) is

y = ax/(-2a) = -x/2 with tangent of -1/2,

The point P on the circle is (-4a, 2a) since it must be equal distance from circle origin (-2a,a) to (0,0 )

Alternatively,

(x+2a)² + (-x/2-a)² = 5a²

x²+ 4ax + 4a² + x²/4 +ax + a²= 5a²

5x²/4 + 5ax = o

x(5/4x + 5a) = 0

x=-4a

when x = -4a,

(-4a+2a)² + (y-a)² = 5a²

(y-a)²=a²

y = 2a

So P is ( -4a, 2a)

the tangent of the straight line to P  has slope of  2 since it is perpendicular  to line y = -x/2

So the equation is y = 2x + c, on the point (-4a,2a)

2a = 2(-4a) + c

c= 10a

y = 2x + 10a

when y = 0 , x= -5a

So the coord. of tangent of OP that meets the x_axis at (-5a,0)

 

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Thanks for this answer.

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