Need help with circle plz

The straight line 3x-4y=4 is a tangent to the circle S at the point (-4,-4). GIven that the centre of the circle lies on the line x+y+7=0, find the equation of the circle.

> Need help with this.TQ

Since centre lies on the line y = -(x+7),

let : C ≡ ( h, -(h+7)) ........................................ (1)

Since the line L : 3x - 4y - 4 = 0 touches

this circle at the point A(-4,-4),

∴ CA = dist. of C from the line L

∴ √[(h+4)²+(-(h+7)+4)²] = | 3(h) - 4(-(h+7)) - 4 | / √(3²+(-4)²)

Now, square , simplify and solve for h.

I'll leave the rest to you now.
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Let the centre be (t,-7-t).
The radius of the circle is the perpendicular distance of this point from the line 3x-4y=4.
R=(3t+4(t+7)-4)/5=(7t+24)/5........(1)
The equation of the circle is (x-t)^2+(y+t+7)^2=R^2.........(2)
Use the value of R from 1 and substitute x=-4,y=-4 to get the value of t. Now use this value of t in eqn. 2 to get the equation of the circle. I haven't checked, but you might get two values of t so there are two circles satisfying the given conditions.