Step 1: Understand the condition for conjugate points.
Two points \((x_1, y_1)\) and \((x_2, y_2)\) are said to be conjugate with respect to a circle \(x^2 + y^2 + 2gx + 2fy + c = 0\) if the polar of one point passes through the other point. The condition for conjugacy is \(x_1x_2 + y_1y_2 + g(x_1+x_2) + f(y_1+y_2) + c = 0\).
Step 2: Identify the circle equation and the given points.
The given circle equation is \(x^2 + y^2 = 25\), which can be written as \(x^2 + y^2 - 25 = 0\). Comparing this to the general form \(x^2 + y^2 + 2gx + 2fy + c = 0\), we have:
\(g = 0\)
\(f = 0\)
\(c = -25\)
The two given points are \((x_1, y_1) = (1, a)\) and \((x_2, y_2) = (b, 2)\).
Step 3: Apply the conjugate points condition.
Substitute the coordinates of the points and the coefficients of the circle into the conjugacy condition:
\(x_1x_2 + y_1y_2 + g(x_1+x_2) + f(y_1+y_2) + c = 0\) \[ (1)(b) + (a)(2) + 0(1+b) + 0(a+2) + (-25) = 0 \] \[ b + 2a - 25 = 0 \] Rearrange the terms to find the relationship between \(a\) and \(b\): \[ 2a + b = 25 \] Step 4: Calculate the required expression \(4a + 2b\).
We need to find the value of \(4a + 2b\).
Notice that \(4a + 2b\) is simply \(2\) times the expression \(2a + b\): \[ 4a + 2b = 2(2a + b) \] Substitute the value of \(2a + b\) from Step 3: \[ 4a + 2b = 2(25) \] \[ 4a + 2b = 50. \]
| List-I | List-II | ||
|---|---|---|---|
| (A) | $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ | (I) | $[\frac{1}{3} , 1 ]$ |
| (B) | $(x)=|[x]|,x \in [R$ | (II) | Z |
| (C) | $h(x) = |x - [x]| , x \in [R$ | (III) | W |
| (D) | $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ | (IV) | [0, 1) |
| (V) | { -1, 1} | ||
| List I | List II | ||
|---|---|---|---|
| (A) | $\lambda=8, \mu \neq 15$ | 1. | Infinitely many solutions |
| (B) | $\lambda \neq 8, \mu \in R$ | 2. | No solution |
| (C) | $\lambda=8, \mu=15$ | 3. | Unique solution |
The equation of a circle which touches the straight lines $x + y = 2$, $x - y = 2$ and also touches the circle $x^2 + y^2 = 1$ is: