x2 + y2 – 3x + y = 0
x2 + y2 + x + 3y = 0
x2 + y2 + 2y – 1 = 0
x2 + y2 + x + y = 0
Given that the mirror image of the orthocenter lies on the circumcircle:
The image of the point (1, 1) reflected over the x-axis is (1, -1). The image of the point (1, 1) reflected over the line \(x+y+1\)=0 is (-2, -2).
Therefore, the circle passing through both (1, -1) and (-2, -2) is determined.
Thus, the circle represented by the equation x2 + y2 + x + 3y = 0 satisfies this condition.
Hence the correct option is B
A positive, singly ionized atom of mass number $ A_M $ is accelerated from rest by the voltage $ 192 \, \text{V} $. Thereafter, it enters a rectangular region of width $ w $ with magnetic field $ \vec{B}_0 = 0.1\hat{k} \, \text{T} $. The ion finally hits a detector at the distance $ x $ below its starting trajectory. Which of the following option(s) is(are) correct?
$ \text{(Given: Mass of neutron/proton = } \frac{5}{3} \times 10^{-27} \, \text{kg, charge of the electron = } 1.6 \times 10^{-19} \, \text{C).} $