Q.If the length of the tangent from $(5, 4)$ to the circle $x^2 + y^2 + 2ky = 0$ is $1$, then find $k$.

Dividing by $2$ and substituting $(4,2)$ gives $S_{11}=4.5>0$, so the point is outside the circle.

Divide the equation by $2$ to standard form:

$x^2+y^2-\dfrac52x-2y-\dfrac32=0,\qquad S\equiv x^2+y^2-\tfrac52x-2y-\tfrac32.$

Evaluate $S_{11}=S(4,2)$:

$S_{11}=4^2+2^2-\tfrac52(4)-2(2)-\tfrac32=16+4-10-4-1.5=4.5.$

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The chord of contact of $(x_1,y_1)$ w.r.t. $x^2+y^2=a^2$ is $xx_1+yy_1=a^2$; here it gives $x+y=9$.

The chord of contact (polar) of a point $(x_1,y_1)$ with respect to the circle $x^2+y^2=a^2$ is obtained by the substitution $x^2 \to xx_1$, $y^2 \to yy_1$: …

Substitute the three known points $(0,0)$, $(a,0)$, $(0,b)$ into the general circle equation and solve for the constants.

Let the circle be $x^2+y^2+2gx+2fy+c=0$. It passes through $(0,0)$, and the intercepts on the axes are $a$ (at $(a,0)$) and $b$ (at $(0,b)$).

Through $(0,0)$: $\;c=0$.

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The normal at $P(3,5)$ passes through the centre $(5,1)$; its equation is $2x + y - 11 = 0$.

For $S \equiv x^2 + y^2 - 10x - 2y + 6 = 0$, the centre is $\left(-\tfrac{-10}{2}, -\tfrac{-2}{2}\right) = (5, 1)$.

A normal to a circle at any point of it always passes through the centre. So the normal at $P(3,5)$ is the line through $(3,5)$ and $(5,1)$.

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Two points $(x_1,y_1)$, $(x_2,y_2)$ are conjugate w.r.t. a circle $S=0$ when $S_{12}=0$, i.e. $x_1x_2+y_1y_2+g(x_1+x_2)+f(y_1+y_2)+c=0$.

For $x^2+y^2=35$, we have $g=f=0$, $c=-35$, so the conjugacy condition reduces to

$x_1x_2+y_1y_2-35=0$

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The chord of contact of an external point $(x_1,y_1)$ w.r.t. circle $S=0$ is $S_1=0$, i.e. $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$.

For $x^2+y^2-5x+4y-2=0$, $g=-\frac{5}{2}$, $f=2$, $c=-2$. With $(x_1,y_1)=(0,5)$:

$x(0)+y(5)+\left(-\frac{5}{2}\right)(x+0)+2(y+5)+(-2)=0$ …

The length of the tangent from $(x_1,y_1)$ to a circle $S=0$ is $\sqrt{S_1}$, where $S_1$ is obtained by substituting $(x_1,y_1)$ into $S$.

Circle: $S \equiv x^2+y^2+2ky=0$, so $g=0,\ f=k,\ c=0$.

Length of tangent from $(5,4)$ $=\sqrt{S_1}$ where …

The polar of $(x_1,y_1)$ with respect to $x^2+y^2=a^2$ is $xx_1+yy_1=a^2$.

Write the circle in standard form: $2x^2+2y^2=11 \Rightarrow x^2+y^2=\frac{11}{2}$, so $a^2=\frac{11}{2}$.

The polar of a point $(x_1,y_1)$ with respect to $x^2+y^2=a^2$ is obtained by replacing $x^2\to xx_1$ and $y^2\to yy_1$:

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The square of the tangent length from a point to a circle $S=0$ equals $S_1$, obtained by substituting the point into the circle's equation.

For circle $S\equiv x^2+y^2+2gx+2fy+c=0$, the square of the tangent length from $(x_1,y_1)$ is $S_1=x_1^2+y_1^2+2gx_1+2fy_1+c$.

Here $S\equiv x^2+y^2-5x+4y+k=0$ and the point is $(2,5)$: …