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 $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
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Important Questions from Algebra
- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.
- The number of $3 \times 2$ matrices A, which can be formed using the elements of the set $\{-2, -1, 0, 1, 2\}$ such that the sum of all the diagonal elements of $A^T A$ is 5, is ______
- The number of numbers greater than 5000, less than 9000 and divisible by 3, that can be formed using the digits 0, 1, 2, 5, 9, if the repetition of the digits is allowed, is ______
- Let $P = [p_{ij}]$ and $Q = [q_{ij}]$ be two square matrices of order 3 such that $q_{ij} = 2^{(i + j - 1)} p_{ij}$ and $\det(Q) = 2^{10}$. Then the value of $\det(\text{adj}(\text{adj } P))$ is:
