1.Find the equation of the circle which passes through the intersection of the circles x2+y2+6x+4y-5=0 and x2+y2-6x-8y+1=0 and whose radius is 3.

2.​Find the equation of the circle which passes through the points intersection of x2+y2-4=0 and x2+y2-2x-4y+4=0 and touches the line x+=0.

2)

The equations of the given intersecting circles are
S = x² +y² -4 = 0, and
S' = x² +y² -2 x -4y +4 = 0
The equation of the common chord of these circles is
S -S' = 0   => l = 2 x +4 +4 y -8 = 0
The equation of the family of circles passing through the intersection of the given circles is
x² +y² -4 +k (2x +4 +4y -8) = 0 ...(i)
Its center is (-k,-2k) and
radius = [k² +4k² +4 +8k] = [5k² +8k +4].
For the particular member of the family which touch the line x +2y = 0, we have
|-k +2 (-2k)|/ [1² +2²] = [5k² +8k +4]
=>   5|k|/5 = [5k² +8k +4]
=> 5 k² = 5 k²+8 k +4
=> 8 k +4 = 0 => k = -1/2
Substituting this value of k in (i), the equation of the required circle is
x² +y² -4 - (1/2)(2x +4y -8) = 0 i.e. x² + y² -x -2y = 0.

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