Step 1: Rewrite the Circles in Standard Form
1. First Circle: \[ x^2 + y^2 - 2x + 4y + 1 = 0 \] Complete the square: \[ (x^2 - 2x) + (y^2 + 4y) = -1 \] \[ (x - 1)^2 + (y + 2)^2 = 1 + 4 - 1 = 4 \] So, the center is \( C_1(1, -2) \) and radius \( r_1 = 2 \).
2. Second Circle: \[ x^2 + y^2 - 4x - 2y + 4 = 0 \] Complete the square: \[ (x^2 - 4x) + (y^2 - 2y) = -4 \] \[ (x - 2)^2 + (y - 1)^2 = 4 + 1 - 4 = 1 \] So, the center is \( C_2(2, 1) \) and radius \( r_2 = 1 \).
Step 2: Find the Distance Between Centers
Calculate the distance \( d \) between \( C_1(1, -2) \) and \( C_2(2, 1) \): \[ d = \sqrt{(2 - 1)^2 + (1 - (-2))^2} = \sqrt{1 + 9} = \sqrt{10} \] Step 3: Determine the Slopes of Direct Common Tangents
For two circles with centers \( C_1(x_1, y_1) \) and \( C_2(x_2, y_2) \), radii \( r_1 \) and \( r_2 \), the slopes \( m \) of the direct common tangents satisfy the condition: \[ \frac{|m(x_2 - x_1) - (y_2 - y_1)|}{\sqrt{m^2 + 1}} = |r_1 - r_2| \] Plugging in the known values: \[ \frac{|m(2 - 1) - (1 - (-2))|}{\sqrt{m^2 + 1}} = |2 - 1| = 1 \] Simplify: \[ \frac{|m - 3|}{\sqrt{m^2 + 1}} = 1 \] Square both sides: \[ (m - 3)^2 = m^2 + 1 \] Expand and solve: \[ m^2 - 6m + 9 = m^2 + 1 \] \[ -6m + 9 = 1 \] \[ -6m = -8 \Rightarrow m = \frac{4}{3} \] Final Answer The slope of one of the direct common tangents is \(\boxed{\dfrac{4}{3}}\).
| 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: