The equation of a tangent to a circle at a given point

Question

18. Show that the equation of the tangent $PT$ at the point $P \left(\frac{1}{5}, \frac{3}{5}\right)$ on the circle

$$x^{2} + y^{2} + 8x + 10y - 8 = 0$$

is $3x + 4y - 3 = 0$.

Find the equations of the chords, each of length $4\ \sqrt{10}$, which are parallel to $PT$.

I proceeded as I've been taught:

The centre of the circle is $(-4, -5)$.

The gradient of the radius to $P$ is $\left.\frac{28}{5}\right/\frac{21}{5} = \frac{4}{3}$.

The gradient the reqd. tangent is $-\frac{3}{4}$.

$\therefore$ The equation of the tangent is:

$y + 5 = -\dfrac{3}{4}(x + 4)$

$4y + 20 = -3x - 12$

$3x + 4y + 32 = 0 \nLeftrightarrow 3x + 4y - 3 = 0$

Is this a misprint in the book, or am I losing my marbles? Since I'm practising in my spare time with nothing more than the book itself, it's hard for me to be sure. But given that I've got the coefficients of $x$ and $y$ right and I can't find an error, I'm forced to conclude that either the problem is malformed, or I have seriously misunderstood something.

Answer 1

Take $$y-\color{green}{\frac{3}{5}}=-\frac{3}{4}\left(x-\color{green}{\frac{1}{5}}\right)$$ instead $$y\color{red}{+5}=-\frac{3}{4}\left(x\color{red}{+4}\right)$$

Answer 2

the slope of the tangent is $-\frac{3}{4}$ thus you have $$y=-\frac{3}{4}x+n$$ setting the coordinates of P in this equation you will find $n=\frac{3}{4}$

Answer 3

the radius at the point $P =(1/5, 3/5)$ has a slope of $\frac43.$ radius of the circle is $5$ and if half the chord is $2\sqrt{10},$ then by phytagora's theorem, it is $$\sqrt{7^2 - (2\sqrt{10})^2} = 3.$$ a point on the diameter with one endpoint $P$ is of the form $$(-4+3k, -5 + 4k) $$ equating the distance from the center $3$, we get $$k = \pm 3/5$$ the equation of the two chords that are of length $4\sqrt{10}$ are $$3x+4y = 3(-4+3k) + 4(-5+4k)=-32+25k= -17, -42. $$