1. Given circle and point

The circle is $x^2+y^2=4,$ so its center is $(0,0)$ and radius is $2$.

The fixed point is $P_0=(1,1).$

We must find the intersection points of tangents at the endpoints of three chords through $P_0$:

• horizontal chord $E_1E_2$
• vertical chord $F_1F_2$
• chord $G_1G_2$ of slope $-1$

Then check which curve contains $E_3,F_3,G_3$.

2. Useful fact: chord of contact / pole relation for the circle

For the circle $x^2+y^2=4,$ the tangent at a point $(x_1,y_1)$ on the circle is $xx_1+yy_1=4.$

If tangents at two points $A,B$ on the circle meet at $T=(h,k)$, then the chord joining the contact points $A,B$ is the polar of $T$, namely $hx+ky=4.$

So for each given chord, if its equation is known, we can identify the tangent-intersection point directly.

3. Chord $E_1E_2$: through $(1,1)$ and parallel to the $x$-axis

A horizontal line through $(1,1)$ is $y=1.$

This is the chord $E_1E_2$.

If tangents at $E_1,E_2$ meet at $E_3=(h,k)$, then its polar is $hx+ky=4.$ Since this polar must be the line $y=1$, we compare: $0\\cdot x + 1\cdot y = 1.$ To write it in the form $hx+ky=4$, multiply by $4$: $0\cdot x + 4y=4.$ Hence $h=0,\quad k=4.$ So $E_3=(0,4).$

4. Chord $F_1F_2$: through $(1,1)$ and parallel to the $y$-axis

A vertical line through $(1,1)$ is $x=1.$

This is the chord $F_1F_2$.

Let tangents at $F_1,F_2$ meet at $F_3=(h,k)$. Its polar is $hx+ky=4.$ Since this must be the line $x=1$, write $x=1 \iff 4x=4.$ Thus $h=4,\quad k=0.$ So $F_3=(4,0).$

5. Chord $G_1G_2$: through $(1,1)$ and slope $-1$

Equation of the line through $(1,1)$ with slope $-1$: $y-1=-1(x-1).$ So $x+y=2.$

Let tangents at $G_1,G_2$ meet at $G_3=(h,k)$. Then its polar is $hx+ky=4.$ Since this must be the line $x+y=2$, multiply by $2$: $2x+2y=4.$ Hence $h=2,\quad k=2.$ So $G_3=(2,2).$

6. Check which curve contains $E_3,F_3,G_3$

We have found: $E_3=(0,4),\quad F_3=(4,0),\quad G_3=(2,2).$

Now test the options.

Option A: $x+y=4$

• For $(0,4)$: $0+4=4$ ✓
• For $(4,0)$: $4+0=4$ ✓
• For $(2,2)$: $2+2=4$ ✓

So all three lie on this line.

Option B: $(x-4)^2+(y-4)^2=16$

• For $(2,2)$: $(2-4)^2+(2-4)^2=4+4=8\ne16$

So false.

Option C: $(x-4)(y-4)=4$

• For $(2,2)$: $(-2)(-2)=4$ ✓
• For $(0,4)$: $(-4)(0)=0\ne4$

So false.

Option D: $xy=4$

• For $(0,4)$: $0\ne4$

So false.

7. Conclusion

The points $E_3,F_3,G_3$ lie on $x+y=4.$ So the correct option is A.

8. Comparison with stored answer

Stored correct answer: A

My derived answer: A

They agree.