First, let's try to convert this equation to the form:
#color(white)(XX)(x-m)^2+(y-n)^2=r^2#
Where:
• (m,n) is the center of the circle
• r is the radius of the circle
We know that #r=6#, so our first step will be to have #6^2# or #36# on the right side.
#[1]color(white)(XX)x^2+y^2-2kx+4y-7=0#
#[2]color(white)(XX)x^2+y^2-2kx+4y-7+36=0+36#
#[3]color(white)(XX)x^2+y^2-2kx+4y+29=36#
Now we will try to get #(y-n)^2# by completing the square.
#[4]color(white)(XX)x^2-2kx+y^2+4y+4+25=36#
#[5]color(white)(XX)x^2-2kx+(y^2+4y+4)+25=36#
#[6]color(white)(XX)x^2-2kx+25+(y+2)^2=36#
We will assume that #x^2-2kx+25# is a perfect square trinomial. It may be written as #(x+5)^2# or #(x-5)^2#. But for now, let's keep it at #(x-k)^2#
#[7]color(white)(XX)(x-k)^2+(y+2)^2=36#
(i)
The center of the circle in terms of #k# is (#k#,-2).
Sorry, but I don't know how to find the radius in terms of #k# (but there's no real reason to do so since it is already stated that #r=6#).
(ii)
In a perfect square trinomial, #c=(b/2)^2#. In the PST #x^2-2kx+25#, #b=-2k# and #c=25#. Solving for #k#:
#c=(b/2)^2#
#25=((-cancel2k)/cancel2)^2#
#25=(-k)^2#
#hArrk=+-5#
