Let w=zz+k_1z+k_2iz+ (1+i),k_1,k_2 R.

Question

Let $w=z\vec{z}+{k}_{1}z+{k}_{2}iz+\lambda (1+i),{k}_{1},{k}_{2}\in R$. Let $Re(w)=0$ be the circle $C$ of radius 1 in the first quadrant touching the line $y=1$ and the $y$-axis. If the curve $Im(w)=0$ intersects $C$ at $A$ and $B$, then $30(AB{)}^{2}$ is equal to \_\_\_\_\_\_\_\_\_\_

Answer: 24

Accepted Answer

Given the expression for $w$ as :

$w=z\vec{z}+{k}_{1}z+{k}_{2}iz+\lambda (1+i),\, {k}_{1},{k}_{2}\in R.$

1. If $w=x+iy$, we can separate this into the real and imaginary parts :

The real part is: $(w)={x}^{2}+{y}^{2}+{k}_{1}x-{k}_{2}y+\lambda =0.$

The imaginary part is: $(w)={k}_{1}y+{k}_{2}x+\lambda =0.$

2. We know the circle C of radius 1 touches the line $y=1$ and the y-axis. The standard form of the equation of a circle is $(x-h{)}^{2}+(y-k{)}^{2}={r}^{2}$, where $(h,k)$ is the center of the circle and $r$ is the radius. If the circle touches y-axis, then the x-coordinate of the center $h$ must be equal to the radius. Also, the circle touches the line $y=1$, so the y-coordinate of the center $k$ must be 1 unit above this line, hence $k=2$. So the circle's equation is $(x-1{)}^{2}+(y-2{)}^{2}=1$.

3. By comparing this with $(w)=0$, we can see that ${k}_{1}=-2$, ${k}_{2}=4$, and $\lambda =4$.

4. The equation for the curve $(w)=0$ becomes $-2y+4x+4=0$, which simplifies to $y=2x+2$.

5. To find the intersection points A and B, we solve the system of equations consisting of the circle and the line. Substituting $y=2x+2$ into the equation of the circle gives $5{x}^{2}-2x=0$, with solutions $x=0$ and $x=2/5$.

6. Substituting $x=0$ and $x=2/5$ into $y=2x+2$ gives $y=2$ and $y=6/5$ respectively. So, the intersection points are A = $(0,2)$ and B = $(2/5,14/5)$.

7. The distance between A and B is $AB=\sqrt{[(2/5{)}^{2}+(4/5{)}^{2}]}=\sqrt{4/5}$.

8. Finally, $30(AB{)}^{2}=30	imes (4/5)=24$.

Solutions · 1