Let the mirror image of a circle c1 :x2 + y2 – 2x – 6y + α = 0 in line y = x + 1 be c2 : 5x2 + 5y2 + 10gx + 10fy + 38 = 0. If r is the radius of circle c2, then α + 6r2 is equal to _________.
The correct answer is 12
c1: x2 + y2 – 2x – 6y + α = 0
Then centre = (1, 3) and
radius \((r)=\sqrt{10−α}\)
Image of (1, 3) w.r.t. line x – y + 1 = 0 is (2, 2)
c2: 5x2 + 5y2 + 10gx + 10fy + 38 = 0
or
\(x^2+y^2+2gx+2fy+\frac{38}{5}=0\)
Then (–g, –f) = (2, 2)
∴ g = f = – 2 …(i)
Radius of \(c_2=r=\sqrt{4+4−\frac{38}{5}}=\sqrt{10−α}\)
\(⇒ \frac{2}{5}=10−α\)
\(∴ α=\frac{48}{5}\) and \(r=\sqrt{\frac{2}{5}}\)
\(∴ α+6r^2=\frac{48}{5}+\frac{12}{5}\)
= 12
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The dimensions of a physical quantity \( \epsilon_0 \frac{d\Phi_E}{dt} \) are similar to [Symbols have their usual meanings]
