If (a, b) is the common point for the circles | vedclass

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(A) The given equations of the circles are:$S_1: x^2+y^2-4x+4y-1=0$$S_2: x^2+y^2+2x-4y+1=0$The common points of the circles lie on the radical axis,gi

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(A) The given equations of the circles are:$S_1: x^2+y^2-4x+4y-1=0$$S_2: x^2+y^2+2x-4y+1=0$The common points of the circles lie on the radical axis,given by $S_1 - S_2 = 0$.$(x^2+y^2-4x+4y-1) - (x^2+y^2+2x-4y+1) = 0$$-6x + 8y - 2 = 0$$3x - 4y + 1 = 0 \implies x = \frac{4y-1}{3}$Substitute $x$ into $S_2=0$:$(\frac{4y-1}{3})^2 + y^2 + 2(\frac{4y-1}{3}) - 4y + 1 = 0$$\frac{16y^2-8y+1}{9} + y^2 + \frac{8y-2}{3} - 4y + 1 = 0$$16y^2-8y+1 + 9y^2 + 24y - 6 - 36y + 9 = 0$$25y^2 - 20y + 4 = 0$$(5y-2)^2 = 0 \implies y = 2/5$Then $x = \frac{4(2/5)-1}{3} = \frac{8/5 - 5/5}{3} = \frac{3/5}{3} = 1/5$Thus,$(a, b) = (1/5, 2/5)$.$a^2+b^2 = (1/5)^2 + (2/5)^2 = 1/25 + 4/25 = 5/25 = 1/5$.

If $(a, b)$ is the common point for the circles $x^2+y^2-4x+4y-1=0$ and $x^2+y^2+2x-4y+1=0$,then $a^2+b^2=$

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