Get the equations of both lines going through $0$ which have a distance of 5 from the point $(1,7)$.
How to handle this problem? We have this formula:
If line $l$ is $ax+by=c$, distance $ P(x,y) $ to line $l$:
$ \dfrac{|ax+by-c|}{\sqrt{a^2+b^2}}$
Get the equations of both lines going through $0$ which have a distance of 5 from the point $(1,7)$.
How to handle this problem? We have this formula:
If line $l$ is $ax+by=c$, distance $ P(x,y) $ to line $l$:
$ \dfrac{|ax+by-c|}{\sqrt{a^2+b^2}}$
Here is another approach, but I think this is more geometric than analytic geometry.
Let $Q(x,y)$ be the leg of the perpendicular from $P$ onto $l$.
Then, $PQ=5$ thus
$(x-1)^2+(y-7)^2=25$
Also, by the Pytagorean Theorem we have
$PQ^2 +QO^2=PO^2$
or
$25+x^2+y^2=50$
The two equations yield
$x^2+y^2=25$ $2x+7y=50$
solving for $x$ in the second equation and plugging in the first gives a quadratic, so the rest is easy.
Find the circle with radius $5$ and center $(1,7)$ and then find the lines passing through the origin and which is a tangent to the circle by using the formula $y = mx+r\cdot\sqrt{m^2+ 1}$ and as it passes through tho origin $r.\sqrt{m^2+1}=0$ so your line now becomes $y=mx$ if distance of this line from $(1,7)$ is $5$, then you get a quadratic in $m$.
Solving this you can find out the values of $m$ and substituting in the line equation, we get the required equation.
First find the equation of the circle of radius 5 around the point. Then the lines you are looking for are the tangents of this circle which pass through 0.
Here the lines pass through the origin $(0,0),$ so $c=0$
$\frac{|a\cdot 1+b \cdot 7|}{\sqrt{a^2+b^2}}=5$
Squaring, $(a+7b)^2=25(a^2+b^2)$
$\implies 24a^2-14ab-24b^2=0$
Solving for $a, a=\frac{4b}3$ or $-\frac{3b}4$
If $a=\frac{4b}3, \frac{4b}3x+by=0\implies 3y+4x=0$ as $b\ne 0$
If $a=-\frac{3b}4, -\frac{3b}4x+by=0\implies -3x+4y=0$ as $b\ne 0$
Here the lines pass through the origin $(0,0),$
the equation of the required line can be $y=mx$ where the gradient $m$ is finite, as the line is not the Y-axis.
So the distance of $(1,7)$ from $mx-y=0$ is $\frac{|m\cdot 1-1\cdot 7|}{\sqrt{m^2+1}}=5$
Squaring we get, $(m-7)^2=25(m^2+1)$
$24m^2+14m-24=0\implies m=-\frac 4 3$ or $\frac 3 4$