If the locus of $z \in C$, such that Re $\left(\frac{z-1}{2z + i}\right)$+Re $\left(\frac{\bar{z}-1}{2\bar{z}-i}\right)$ = 2, is a circle of radius r and center (a, b), then $\frac{15ab}{r^2}$ is equal to :
Simplifying the Locus Equation
The given equation is: Re $\left(\frac{z-1}{2z + i}\right)$ + Re $\left(\frac{\bar{z}-1}{2\bar{z}-i}\right)$ = 2
Let $w = \frac{z-1}{2z + i}$. We know that $\overline{w} = \frac{\overline{z-1}}{\overline{2z+i}} = \frac{\bar{z}-1}{2\bar{z}-i}$.
The equation simplifies using the property Re($w$) = Re($\overline{w}$): Re($w$) + Re($w$) = 2 $2 \times$ Re($w$) = 2 Re($w$) = 1
Therefore, the condition reduces to finding the locus where Re $\left(\frac{z-1}{2z + i}\right)$ = 1.
Finding the Circle Equation
Let $z = x + iy$. Substitute into the expression:
$z-1 = (x-1) + iy$
$2z+i = 2(x+iy) + i = 2x + i(2y+1)$
Calculate the fraction $\frac{z-1}{2z+i}$:
$\frac{(x-1) + iy}{2x + i(2y+1)} = \frac{((x-1) + iy)(2x - i(2y+1))}{(2x + i(2y+1))(2x - i(2y+1))}$
Numerator = $(x-1)(2x) + y(2y+1) + i(y(2x) - (x-1)(2y+1))$
Numerator = $(2x^2 - 2x + 2y^2 + y) + i(2xy - (2xy + x - 2y - 1))$
Numerator = $(2x^2 - 2x + 2y^2 + y) + i(-x + 2y + 1)$
Denominator = $(2x)^2 + (2y+1)^2 = 4x^2 + 4y^2 + 4y + 1$
The real part is Re $\left(\frac{z-1}{2z+i}\right) = \frac{2x^2 - 2x + 2y^2 + y}{4x^2 + 4y^2 + 4y + 1}$.
Set the real part equal to 1:
$\frac{2x^2 - 2x + 2y^2 + y}{4x^2 + 4y^2 + 4y + 1} = 1$
$2x^2 - 2x + 2y^2 + y = 4x^2 + 4y^2 + 4y + 1$
Rearrange terms:
$2x^2 + 2y^2 + 2x + 3y + 1 = 0$
Divide by 2 to get the standard form:
$x^2 + y^2 + x + \frac{3}{2}y + \frac{1}{2} = 0$
Identifying Circle Parameters (Center and Radius)
Compare $x^2 + y^2 + x + \frac{3}{2}y + \frac{1}{2} = 0$ with the general circle equation $x^2 + y^2 + 2gx + 2fy + c = 0$.
- Center $(a, b) = (-g, -f)$.
- From the equation, $2g = 1 \implies g = \frac{1}{2}$ and $2f = \frac{3}{2} \implies f = \frac{3}{4}$.
- Thus, the center is $(a, b) = (-\frac{1}{2}, -\frac{3}{4})$.
- The radius squared $r^2 = g^2 + f^2 - c$.
- Here, $c = \frac{1}{2}$.
- $r^2 = (\frac{1}{2})^2 + (\frac{3}{4})^2 - \frac{1}{2} = \frac{1}{4} + \frac{9}{16} - \frac{8}{16} = \frac{4+9-8}{16} = \frac{5}{16}$.
Calculating the Final Expression
We need to find the value of $\frac{15ab}{r^2}$.
- $a = -\frac{1}{2}$
- $b = -\frac{3}{4}$
- $r^2 = \frac{5}{16}$
- $ab = (-\frac{1}{2}) \times (-\frac{3}{4}) = \frac{3}{8}$
Substitute these values into the expression:
$\frac{15ab}{r^2} = \frac{15 \times (\frac{3}{8})}{\frac{5}{16}} = \frac{\frac{45}{8}}{\frac{5}{16}}$
$\frac{45}{8} \times \frac{16}{5} = \frac{45}{5} \times \frac{16}{8} = 9 \times 2 = 18$.
Similar Questions
Let A = {-3, -2, -1, 0, 1, 2, 3}. Let R be a relation on A defined by xRy if and only if $0\le x^2+2y\le 4$. Let $l$ be the number of elements in R and $m$ be the minimum number of elements required to be added in R to make it a reflexive relation. Then $l + m$ is equal to
Let $\alpha$ and $\beta$ be the roots of $x^2 + \sqrt{3}x-16=0$, and $\gamma$ and $\delta$ be the roots of $x^2 + 3x - 1 = 0$. If $P_n = \alpha^n + \beta^n$ and $Q_n = \gamma^n + \delta^n$, then $\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} + \frac{Q_{25}-Q_{23}}{Q_{24}}$ is equal to
Let the domain of the function $f (x) = \log_2 \log_4 \log_6(3 + 4x -x^2)$ be $(a, b)$.
If $\int_{b-a}^{b+a}[x^2] dx=p-\sqrt{q}-\sqrt{r}$, $p,q,r\in N$, $gcd(p,q,r) = 1$, where $[.]$ is the greatest integer function, then $p + q + r$ is equal to
Let A be a matrix of order $3 \times 3$ and $|A| = 5$. If $|2\text{adj} (3A \text{adj} (2A))| = 2^\alpha \cdot 3^\beta \cdot 5^\gamma$, $\alpha, \beta, \gamma \in N$, then $\alpha + \beta + \gamma$ is equal to
Let $a_1, a_2, a_3,....$ be a G.P. of increasing positive numbers. If $a_3a_5 = 729$ and $a_2 + a_4 = \frac{111}{4}$, then $24 (a_1 + a_2 + a_3)$ is equal to
All five letter words are made using all the letters A, B, C, D, E and arranged as in an English dictionary with serial numbers. Let the word at serial number $n$ be denoted by $W_n$. Let the probability $P(W_n)$ of choosing the word $W_n$ satisfy $P(W_n) = 2P(W_{n-1})$, $n > 1$.
If $P(CDBEA) = \frac{2^\alpha}{2^\beta-1}$, $\alpha, \beta\in N$, then $\alpha + \beta$ is equal to :
- The largest $n \in N$ such that $3^n$ divides $50!$ is :
- The number of sequences of ten terms, whose terms are either 0 or 1 or 2, that contain exactly five 1s and exactly three 2s, is equal to :
- Let A be the set of all functions $f: Z \to Z$ and R be a relation on A such that $R = \{(f, g) : f(0) = g(1) \text{ and } f(1) = g(0)\}$. Then R is :
Let $a \in \mathbf{R}$ and $A$ be a matrix of order $3 \times 3$ such that $\det (A) = -4$ and $A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix}$, where $I$ is the identity matrix of order $3 \times 3$. If $\det ((a+1)\text{adj}((a-1)A))$ is $2^m 3^n$, $m, n \in \{0, 1, 2, \dots, 20\}$, then $m+n$ is equal to :
Important Questions from Algebra
Let A = {-3, -2, -1, 0, 1, 2, 3}. Let R be a relation on A defined by xRy if and only if $0\le x^2+2y\le 4$. Let $l$ be the number of elements in R and $m$ be the minimum number of elements required to be added in R to make it a reflexive relation. Then $l + m$ is equal to
Let $\alpha$ and $\beta$ be the roots of $x^2 + \sqrt{3}x-16=0$, and $\gamma$ and $\delta$ be the roots of $x^2 + 3x - 1 = 0$. If $P_n = \alpha^n + \beta^n$ and $Q_n = \gamma^n + \delta^n$, then $\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} + \frac{Q_{25}-Q_{23}}{Q_{24}}$ is equal to
Let the domain of the function $f (x) = \log_2 \log_4 \log_6(3 + 4x -x^2)$ be $(a, b)$.
If $\int_{b-a}^{b+a}[x^2] dx=p-\sqrt{q}-\sqrt{r}$, $p,q,r\in N$, $gcd(p,q,r) = 1$, where $[.]$ is the greatest integer function, then $p + q + r$ is equal to
Let A be a matrix of order $3 \times 3$ and $|A| = 5$. If $|2\text{adj} (3A \text{adj} (2A))| = 2^\alpha \cdot 3^\beta \cdot 5^\gamma$, $\alpha, \beta, \gamma \in N$, then $\alpha + \beta + \gamma$ is equal to
Let $a_1, a_2, a_3,....$ be a G.P. of increasing positive numbers. If $a_3a_5 = 729$ and $a_2 + a_4 = \frac{111}{4}$, then $24 (a_1 + a_2 + a_3)$ is equal to
