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

I don't know the solution for this question

Chinnapparaj R
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Aakriti Kumal
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3 Answers3

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The slope of your tangent is given by $$2x+2y\cdot y'-2-4y'=0$$ so $$y'(2y-4)=2-2x$$ Can you proceed?

Dr. Sonnhard Graubner
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Hint:

The normal to a circle with centre $(\omega_0, \omega_1)$ at a point $(x_0, x_1)$ is just the line through this point and the centre, so its equation is given by $$\frac{y-\omega_1}{x-\omega_0}=\frac{x_1-\omega_1}{x_0-\omega_0}.$$ Can you find the coordinates of the centre? As to the tangent, it is the perpendicular to the normal passing through $(x_0, x_1)$. Note its slope is $\;-\frac{x_0-\omega_0}{x_1-\omega_1}$, so that an equation is $$y=-\frac{x_0-\omega_0}{x_1-\omega_1}(x-x_0)+x_1.$$

Bernard
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    In the last equation.... instead of x_1 you should type y_1. – Supriyo Banerjee Dec 20 '18 at 16:11
  • I realise my notations aren't very good, but it has indeed to be $x_1$. The problem comes from the fact I chose $\omega$ for the name of the centre, and there's no letter after $\omega$ to denote the 2nd coordinate of this centre, so I had to use a system of indices. – Bernard Dec 20 '18 at 16:23
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Equation of the tangent is $2x + 3y -(x+2)-2(y+3)+3=0$ i.e $x+y-5=0$

Therefore the slope of the tangent is $-1$ and hence the slope of the normal of the circle at $(2,3)$ is $1$.

Therefore the equation of the normal at $(2,3)$ is.

$\frac{y-3}{x-2} =1$

i.e $x-y+1=0$

Supriyo Banerjee
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