Let z=x+iy so that
2∣x+i(y+3)∣=∣x+i(y−1)∣
⇒2x2+(y+3)2=x2+(y−1)2
⇒4x2+4(y+3)2=x2+(y−1)2
⇒3x2+3y2+26y+35=0
⇒x2+y2+326y+335=0
⇒r=0+9169−335
⇒r=964=38
JEE Main 2017 — Mathematics Algebra
Let z∈C, the set of complex numbers. Then the equation, 2∣z+3i∣−∣z−i∣=0 represents:
Held on 8 Apr 2017 · Verified 6 Jul 2026.
A circle with radius 38
An ellipse with length of minor axis 916
An ellipse with length of major axis 316
A circle with diameter 310
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
If the quadratic equation $(\lambda+2)x^2-3\lambda x+4\lambda=0$, $\lambda \neq -2$, has two positive roots, then the number of possible integral values of $\lambda$ is:
The value of $\frac{{ }^{100} \mathrm{C}_{50}}{51}+\frac{{ }^{100} \mathrm{C}_{51}}{52}+\ldots.+\frac{{ }^{100} \mathrm{C}_{100}}{101}$ is :
Let $e_1$ and $e_2$ be two distinct roots of the equation $x^2 - ax + 2 = 0$. Let the sets $\{a \in \mathbb{R} : e_1 \text{ and } e_2 \text{ are the eccentricities of hyperbolas}\} = (\alpha, \beta)$, and $\{a \in \mathbb{R} : e_1 \text{ and } e_2 \text{ are the eccentricities of an ellipse and a hyperbola, respectively}\} = (\gamma, \infty)$. Then $\alpha^2 + \beta^2 + \gamma^2$ is equal to:
Among the statements : I: If $\left|\begin{array}{ccc}1 & \cos \alpha & \cos \beta \\ \cos \alpha & 1 & \cos \gamma \\ \cos \beta & \cos \gamma & 1\end{array}\right|=\left|\begin{array}{ccc}0 & \cos \alpha & \cos \beta \\ \cos \alpha & 0 & \cos \gamma \\ \cos \beta & \cos \gamma & 0\end{array}\right|$, then $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=\frac{3}{2}$, and II : If $\left|\begin{array}{ccc}x^{2}+x & x+1 & x-2 \\ 2 x^{2}+3 x-1 & 3 x & 3 x-3 \\ x^{2}+2 x+3 & 2 x-1 & 2 x-1\end{array}\right|=\mathrm{p} x+\mathrm{q}$, then $\mathrm{p}^{2}=196 \mathrm{q}^{2}$,
If $\left(\frac{1}{{ }^{15} \mathrm{C}_{0}}+\frac{1}{{ }^{15} \mathrm{C}_{1}}\right)\left(\frac{1}{{ }^{15} \mathrm{C}_{1}}+\frac{1}{{ }^{15} \mathrm{C}_{2}}\right) \cdots\left(\frac{1}{{ }^{15} \mathrm{C}_{12}}+\frac{1}{{ }^{15} \mathrm{C}_{13}}\right)=\frac{\alpha^{13}}{{ }^{14} \mathrm{C}_{0}{ }^{14} \mathrm{C}_{1} \cdots{ }^{14} \mathrm{C}_{12}}$, then $30 \alpha$ is equal to $\_\_\_\_$.
Work through every JEE Main Algebra PYQ, year by year.