Let $C$ be the circle concentric with the circle, $2x^2+2y^2-6x-10y=183$ and having area $(1/10)th$ of the area of this circle. Then a tangent to $C$, parallel to the line, $3x+y=0$ makes an intercept on the $y$-axis, which is equal to
The given circle is $x^2+y^2-3x-5y-\frac{183}{2}=0$. So, the center is $(\frac32,\frac52)$. And the radius is $10$. Let the radius of $C$ be $r$, so, $$\pi r^2=\frac1{10}\pi(10)^2\implies r^2=10$$ So, the equation of $C$ becomes $$(x-\frac32)^2+(y-\frac52)^2=10$$
Taking derivative, $$\frac{dy}{dx}=\frac{x-\frac32}{3(y-\frac52)}=-3$$
Putting the value of $(x-\frac32)$ in the equation of $C$, I get $y=\frac{215}{82}$. From that, I get $x=\frac{33}{82}$. Thus, the equation of the tangent becomes $$y-\frac{215}{82}=-3(x-\frac{33}{82})$$
Putting $x=0$, I get the $y-$intercept as $\frac{314}{82}$. But the answer is given as $17$.
What is my mistake? Also, is there a quicker way to solve this? Thanks.
For a faster method, observe that the slope of the circle’s radius at the point of tangency is $\frac13$, so the two points of tangency are $$\left(\frac32,\frac52\right)\pm \sqrt{10}\left(\frac3{\sqrt{10}},\frac1{\sqrt{10}}\right) = \left(\frac32\pm3,\frac52\pm1\right).$$ The two tangent lines are therefore $$\left(y-\frac52\pm1\right) = -3\left(x-\frac32\pm3\right)$$ with $y$-intercepts of $-3$ and $17$.
As for where your calculations went wrong, it looks like you made a sign error when you computed the derivative. From the original equation of the circle, we have $$4x+4yy'-6-10y' = (4y-10)y'+(4x-6) = 0,$$ so $$y' = -{2x-3\over2y-5}$$ and $$x-\frac32 = \frac32(2y-5) = 3y-\frac{15}2.$$ You probably forgot to negate the terms that don’t involve $y'$ when you moved them to the other side of the equation while solving for $y'$.