Pregunta
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Find the equation of the circle with a center at \( (0,-4) \) and a radius of? \( x^{2}+(y+4)^{2}=7 \) \( x^{2}-(y+4)^{2}=7 \) \( x^{2}+(y+4)^{2}=40 \) \( x^{2}+(y-4)^{2}=40 \) QuEstion Given the equation \( x=0 \), which equation of a circle makes the equation a tangent to the same circle? \( (x+2)^{2}+(y+3)^{2}=0 \) \( (x+2)^{2}+(y-4)^{2}=9 \) \( (x-3)^{2}+(y+1)^{2}=4 \) \( (x-1)^{2}+(y-2)^{2}=16 \)

Ask by Wright Ward. in the United States
Jan 24,2025

Solución de tutoría real

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The circle with center \( (0, -4) \) and radius \( \sqrt{7} \) is \( x^{2}+(y+4)^{2}=7 \). The circle tangent to the line \( x = 0 \) is \( (x-1)^{2}+(y-2)^{2}=16 \).

Solución

Let's solve the first part of the problem regarding the equation of the circle. ### Part 1: Finding the Equation of the Circle 1. **Known Conditions**: - Center: \( (0, -4) \) - Radius: \( r \) (not specified yet) The standard form of the equation of a circle with center \( (h, k) \) and radius \( r \) is given by: \[ (x - h)^2 + (y - k)^2 = r^2 \] Substituting the center \( (0, -4) \) into the equation, we have: \[ (x - 0)^2 + (y + 4)^2 = r^2 \] This simplifies to: \[ x^2 + (y + 4)^2 = r^2 \] Now, we need to match this with the given options. The options are: 1. \( x^{2}+(y+4)^{2}=7 \) 2. \( x^{2}-(y+4)^{2}=7 \) 3. \( x^{2}+(y+4)^{2}=40 \) 4. \( x^{2}+(y-4)^{2}=40 \) From the first and third options, we can see that they are in the form \( x^2 + (y + 4)^2 = r^2 \). - For option 1, \( r^2 = 7 \) - For option 3, \( r^2 = 40 \) Since both options 1 and 3 are valid equations of circles with the given center, we can conclude that the equations of the circle could be either: - \( x^{2}+(y+4)^{2}=7 \) (if \( r = \sqrt{7} \)) - \( x^{2}+(y+4)^{2}=40 \) (if \( r = \sqrt{40} \)) ### Part 2: Finding the Tangent Circle Next, we need to determine which of the given circles is tangent to the line \( x = 0 \). 1. **Known Condition**: - The line \( x = 0 \) is a vertical line. To be tangent to the line \( x = 0 \), the distance from the center of the circle to the line must equal the radius of the circle. Now, let's analyze each option: 1. \( (x+2)^{2}+(y+3)^{2}=0 \) - Center: \( (-2, -3) \), Radius: \( 0 \) - Distance to line \( x = 0 \): \( 2 \) (not tangent) 2. \( (x+2)^{2}+(y-4)^{2}=9 \) - Center: \( (-2, 4) \), Radius: \( 3 \) - Distance to line \( x = 0 \): \( 2 \) (not tangent) 3. \( (x-3)^{2}+(y+1)^{2}=4 \) - Center: \( (3, -1) \), Radius: \( 2 \) - Distance to line \( x = 0 \): \( 3 \) (not tangent) 4. \( (x-1)^{2}+(y-2)^{2}=16 \) - Center: \( (1, 2) \), Radius: \( 4 \) - Distance to line \( x = 0 \): \( 1 \) (tangent) ### Conclusion - The equations of the circle with center \( (0, -4) \) can be either: - \( x^{2}+(y+4)^{2}=7 \) - \( x^{2}+(y+4)^{2}=40 \) - The circle that is tangent to the line \( x = 0 \) is: - \( (x-1)^{2}+(y-2)^{2}=16 \)

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The Deep Dive

The equation for a circle is traditionally given in the form \( (x - h)^{2} + (y - k)^{2} = r^{2} \), where \( (h,k) \) is the center and \( r \) is the radius. For a center at \( (0,-4) \) and a radius of \( \sqrt{7} \), the equation simplifies to \( x^{2}+(y+4)^{2}=7 \). For the second part regarding the tangent, the tangent line \( x=0 \) indicates that the circle's center must lie on the line \( x=0 \) or directly adjacent to it. The equation \( (x+2)^{2}+(y-4)^{2}=9 \) has a center at \( (-2, 4) \) which is at a distance of exactly 5 from the line \( x=0 \), where a tangent would touch the circle at exactly one point.

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