The equation of a circle which touches the line $y = x$ at $(1, 1)$ and having $y = x - 3$ as a normal, is

4 x^{2} + 4 y^{2} - 20 x + 4 y + 8 = 0

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x^{2} + y^{2} - 2 x + 4 y + 8 = 0

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x^{2} + y^{2} - 10 x - 4 y + 8 = 0

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none of these

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Solution

The correct option is D $4 x^{2} + 4 y^{2} - 20 x + 4 y + 8 = 0$
The equation of the family of circles is
$(x - 1)^{2} + (y - 1)^{2} + λ (x - y) = 0 . . . (i)$
Its centre lies on the line $y = x - 3$ .
$∴ 1 + \frac{λ}{2} = 1 - \frac{λ}{2} - 3 \Rightarrow λ = - 3$
Putting $λ = - 3$ in $(i)$ , we obtain
$x^{2} + y^{2} - 5 x + y + 2 = 0$

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The equation of a circle which touches the line y=x at (1,1) and having y=x−3 as a normal, is

The correct option is D 4x2+4y2−20x+4y+8=0The equation of the family of circles is (x−1)2+(y−1)2+λ(x−y)=0...(i)Its centre lies on the line y=x−3.∴1+λ2=1−λ2−3⇒λ=−3Putting λ=−3 in (i), we obtainx2+y2−5x+y+2=0

The equation of a circle which touches the line $y = x$ at $(1, 1)$ and having $y = x - 3$ as a normal, is