Find the equation of the circle passing through the point (2, 1) and touching the line x + 2y - 1 = 0 at the point (3, -1)

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Solution

Equation of circle is $(x - 3)^{2} + (y + 1)^{2} + λ (x + 2 y - 1) = 0$
Since it passes through the point $(2, 1), 1 + 4 +$ $λ$ (2 + 2 - 1) = 0 $\Rightarrow λ = - \frac{5}{3}$
$∴ c i r c l e i s (x - 3)^{2} + (y + 1)^{2} - \frac{5}{3} (x + 2 y - 1) = 0$
$\Rightarrow 3 x^{2} + 3 y^{2} - 23 x - 4 y + 35 = 0$

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Find the equation of the circle passing through the point (2, 1) and touching the line x + 2y - 1 = 0 at the point (3, -1)

Equation of circle is (x−3)2+(y+1)2+λ(x+2y−1)=0Since it passes through the point (2,1),1+4+ λ (2 + 2 - 1) = 0 ⇒λ=−53∴circleis(x−3)2+(y+1)2−53(x+2y−1)=0⇒3x2+3y2−23x−4y+35=0