Let all the points on the curve $x^2+y^2-10x=0$ are reflected about the line $y=x+3$. If the locus of the reflected points is in the form $x^2+y^2+gx+fy+c=0$, then the value of $(g+f+c)$ is
(A) 38
(B) -28
(C) 28
(D) -38

Question

Accepted Answer

> [!TLDR]
> Reflect the centre $(5,0)$ of $x^2+y^2-10x=0$ about $y=x+3$, keep the same radius, then read off $g,f,c$.

The circle $x^2+y^2-10x=0$ has centre $C=(5,0)$ and radius $r=5$ (completing the square: $(x-5)^2+y^2=25$).

Reflecting a circle about a line simply reflects its centre and keeps the radius unchanged (every point on the circle maps to a point at the same distance $r$ from the reflected centre).

Line: $y=x+3 \Rightarrow x-y+3=0$, i.e. $a=1,\ b=-1,\ c=3$.

Reflection formula for point $(x_0,y_0)$ about $ax+by+c=0$:
$$x'=x_0-\frac{2a(ax_0+by_0+c)}{a^2+b^2},\qquad y'=y_0-\frac{2b(ax_0+by_0+c)}{a^2+b^2}$$

For $(x_0,y_0)=(5,0)$: $ax_0+by_0+c = 5-0+3=8$, and $a^2+b^2=2$, so $d=8/2=4$.

$$x' = 5-2(1)(4) = -3,\qquad y'=0-2(-1)(4)=8$$

So the reflected centre is $(-3,8)$, radius still $5$.

Equation of reflected circle:
$$(x+3)^2+(y-8)^2=25$$
$$x^2+6x+9+y^2-16y+64=25$$
$$x^2+y^2+6x-16y+48=0$$

Comparing with $x^2+y^2+gx+fy+c=0$: $g=6,\ f=-16,\ c=48$.

$$g+f+c = 6-16+48 = 38$$

(Check: point $(0,0)$ lies on the original circle; its reflection is $(-3,3)$, and indeed $(-3+3)^2+(3-8)^2=25$ — confirms the derived circle.)

> [!ANSWER]
> $g+f+c=38$

ANSWER: A

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