If the line $2 x - y + 1 = 0$ touches the circle at the point $(2, 5)$ and the centre of the circle on the line $x + y - 9 = 0$ . Find the equation of the circle.

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Solution

$\begin{matrix} {(x - 2)}^{2} + {(y - 5)}^{2} + λ [2 x - y + 1] = 0 x^{2} + y^{2} + 29 - 4 x - 10 y + λ (2 x - y + 1) = 0 x^{2} + y^{2} + x (2 λ - 4) + y (- λ - 10) + (29 + λ) = 0 C e n t r e : (2 - λ, 5 + \frac{λ}{2}) 2 - λ + 5 + \frac{λ}{2} - 9 = 0 - 2 - \frac{λ}{2} = 0 \frac{λ}{2} = - 2 ∴ λ = - 4 H e n c e, E q u a t i o n o f c i r c l e : x^{2} + y^{2} - 12 x - 6 y + 25 = 0 \end{matrix}$

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If the line 2x−y+1=0 touches the circle at the point (2,5) and the centre of the circle on the line x+y−9=0. Find the equation of the circle.

(x−2)2+(y−5)2+λ[2x−y+1]=0x2+y2+29−4x−10y+λ(2x−y+1)=0x2+y2+x(2λ−4)+y(−λ−10)+(29+λ)=0Centre:(2−λ,5+λ2)2−λ+5+λ2−9=0−2−λ2=0λ2=−2∴λ=−4Hence,Equationofcircle:x2+y2−12x−6y+25=0

If the line $2 x - y + 1 = 0$ touches the circle at the point $(2, 5)$ and the centre of the circle on the line $x + y - 9 = 0$ . Find the equation of the circle.