The number of points of intersection of | z -(4+3 i )|=2 and | z |+| z -4|=6, z ∈ C is :

Question

The number of points of intersection of | z -(4+3 i )|=2 and | z |+| z -4|=6, z ∈ C is : (A) 0 (B) 1 (C) 2 (D) 3

Tardigrade Admin · Accepted Answer

C:(x-4)2+(y-3)2=4 E: ((x-2)2/9)+(y2/5)=1 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/1607b9a0ca0dcdce871aca55d59496ad-.png" /> Lower Extremity of vertical diameter of circle arrow(4,1) Put in ellipse ⇒ ((4-2)2/9)+(1/5)-1 =(4/9)+(1/5)-1 =(29/45)-1 < 0 Two Solution

Q. The number of points of intersection of $∣ z - (4 + 3 i) ∣ = 2$ and $∣ z ∣ + ∣ z - 4∣ = 6, z \in C$ is :

0

1

2

3

$C : (x - 4)^{2} + (y - 3)^{2} = 4$
$E : \frac{( x - 2 ) ^{2}}{9} + \frac{y ^{2}}{5} = 1$

Lower Extremity of vertical diameter of circle $\to (4, 1)$
Put in ellipse $\Rightarrow \frac{( 4 - 2 ) ^{2}}{9} + \frac{1}{5} - 1$
$= \frac{4}{9} + \frac{1}{5} - 1$
$= \frac{29}{45} - 1 < 0$
Two Solution

Q. The number of points of intersection of ∣z−(4+3i)∣=2 and ∣z∣+∣z−4∣=6,z∈C is :

0

1

2

3

C:(x−4)2+(y−3)2=4 E:9(x−2)2​+5y2​=1 Lower Extremity of vertical diameter of circle →(4,1) Put in ellipse ⇒9(4−2)2​+51​−1 =94​+51​−1 =4529​−1<0 Two Solution

Q. The number of points of intersection of $∣ z - (4 + 3 i) ∣ = 2$ and $∣ z ∣ + ∣ z - 4∣ = 6, z \in C$ is :

$C : (x - 4)^{2} + (y - 3)^{2} = 4$
$E : \frac{( x - 2 ) ^{2}}{9} + \frac{y ^{2}}{5} = 1$

Lower Extremity of vertical diameter of circle $\to (4, 1)$
Put in ellipse $\Rightarrow \frac{( 4 - 2 ) ^{2}}{9} + \frac{1}{5} - 1$
$= \frac{4}{9} + \frac{1}{5} - 1$
$= \frac{29}{45} - 1 < 0$
Two Solution