Find the equation of the normal to the circle $x^{2} + y^{2} - 2 x - 4 y + 3 = 0$ at the point $(2, 3)$

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Solution

Equation of normal is : $y - y_{1} = \frac{y_{1} + f}{x_{1} + g} (x - x_{1})$

Equation of circle: $x^{2} + y^{2} + 2 g x + 2 y f + c = 0$

$x^{2} + y^{2} - 2 x - 4 y + 3 = 0$

$f = - 2, g = - 1$

Equation of normal: $(y - 3) = (\frac{3 - 2}{2 - 1}) (x - 2)$ at $(2, 3)$

$\Rightarrow y - 3 = 1 (x - 2)$

$\Rightarrow x - y + 1 = 0$

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Find the equation of the normal to the circle x2+y2−2x−4y+3=0 at the point (2,3)

Equation of normal is : y−y1=y1+fx1+g(x−x1)Equation of circle: x2+y2+2gx+2yf+c=0x2+y2−2x−4y+3=0f=−2,g=−1Equation of normal: (y−3)=(3−22−1)(x−2) at (2,3)⇒y−3=1(x−2)⇒x−y+1=0

Find the equation of the normal to the circle $x^{2} + y^{2} - 2 x - 4 y + 3 = 0$ at the point $(2, 3)$