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Sagot :
Let's solve this problem step by step.
### Part (a): Show that the line does not intersect the circle.
We start by analyzing the equations of the line and the circle:
- The line is given by [tex]\( y = x - 5 \)[/tex].
- The circle is given by [tex]\( (x - 3)^2 + (y - 4)^2 = 4 \)[/tex].
To check if they intersect, we substitute the equation of the line into the equation of the circle. Substitute [tex]\( y \)[/tex] from the line equation into the circle equation:
[tex]\[ (x - 3)^2 + ((x - 5) - 4)^2 = 4 \][/tex]
Simplifying the equation, we have:
[tex]\[ (x - 3)^2 + (x - 9)^2 = 4 \][/tex]
Calculating the squares:
[tex]\[ (x - 3)^2 + (x - 9)^2 = (x^2 - 6x + 9) + (x^2 - 18x + 81) \][/tex]
Combine like terms:
[tex]\[ 2x^2 - 24x + 90 = 4 \][/tex]
Subtract 4 from both sides:
[tex]\[ 2x^2 - 24x + 86 = 0 \][/tex]
Simplify by dividing everything by 2:
[tex]\[ x^2 - 12x + 43 = 0 \][/tex]
To determine if real solutions exist, calculate the discriminant [tex]\( \Delta \)[/tex]:
[tex]\[ \Delta = b^2 - 4ac = (-12)^2 - 4(1)(43) = 144 - 172 = -28 \][/tex]
Since the discriminant is negative ([tex]\(-28\)[/tex]), the quadratic equation [tex]\( x^2 - 12x + 43 = 0 \)[/tex] has no real solutions. Therefore, the line and the circle do not intersect.
### Part (b): Find the coordinates of [tex]\( P \)[/tex] and the exact value of the shortest distance from [tex]\( P \)[/tex] to the circle.
To find the point [tex]\( P \)[/tex] on the line [tex]\( y = x - 5 \)[/tex] that is closest to the center of the circle (3, 4), we minimize the distance between a point on the line and the center of the circle.
Consider a point [tex]\( P \)[/tex] on the line given by the coordinates [tex]\( (x, y) \)[/tex] where [tex]\( y = x - 5 \)[/tex]. Let [tex]\( P = (x, x - 5) \)[/tex].
The distance [tex]\( D \)[/tex] between the point [tex]\( P \)[/tex] and the center of the circle [tex]\( (3, 4) \)[/tex] is given by the distance formula:
[tex]\[ D = \sqrt{(x - 3)^2 + ((x - 5) - 4)^2} \][/tex]
Simplify the expression inside the square root:
[tex]\[ D = \sqrt{(x - 3)^2 + (x - 9)^2} \][/tex]
To minimize this distance, we can consider the derivative. However, we can also directly use symmetry and geometric considerations. Since the equation of the line is linear and we're calculating a distance to a fixed point, the closest [tex]\( x \)[/tex] will be the perpendicular projection.
Minimize the distance, recognize that it happens when the derivative of [tex]\( D \)[/tex] with respect to [tex]\( x \)[/tex] is zero, find the critical points.
Solving the differentiation results:
[tex]\[ P = (6, 1) \][/tex]
Finally, calculate the exact distance from [tex]\( P = (6, 1) \)[/tex] to the center of the circle [tex]\( (3, 4) \)[/tex]:
[tex]\[ D = \sqrt{(6 - 3)^2 + (1 - 4)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \][/tex]
### Summary of Results:
(a) The line does not intersect the circle.
(b) The coordinates of [tex]\( P \)[/tex] are [tex]\( (6, 1) \)[/tex] and the exact value of the shortest distance from [tex]\( P \)[/tex] to the circle is [tex]\( 3\sqrt{2} \)[/tex].
### Part (a): Show that the line does not intersect the circle.
We start by analyzing the equations of the line and the circle:
- The line is given by [tex]\( y = x - 5 \)[/tex].
- The circle is given by [tex]\( (x - 3)^2 + (y - 4)^2 = 4 \)[/tex].
To check if they intersect, we substitute the equation of the line into the equation of the circle. Substitute [tex]\( y \)[/tex] from the line equation into the circle equation:
[tex]\[ (x - 3)^2 + ((x - 5) - 4)^2 = 4 \][/tex]
Simplifying the equation, we have:
[tex]\[ (x - 3)^2 + (x - 9)^2 = 4 \][/tex]
Calculating the squares:
[tex]\[ (x - 3)^2 + (x - 9)^2 = (x^2 - 6x + 9) + (x^2 - 18x + 81) \][/tex]
Combine like terms:
[tex]\[ 2x^2 - 24x + 90 = 4 \][/tex]
Subtract 4 from both sides:
[tex]\[ 2x^2 - 24x + 86 = 0 \][/tex]
Simplify by dividing everything by 2:
[tex]\[ x^2 - 12x + 43 = 0 \][/tex]
To determine if real solutions exist, calculate the discriminant [tex]\( \Delta \)[/tex]:
[tex]\[ \Delta = b^2 - 4ac = (-12)^2 - 4(1)(43) = 144 - 172 = -28 \][/tex]
Since the discriminant is negative ([tex]\(-28\)[/tex]), the quadratic equation [tex]\( x^2 - 12x + 43 = 0 \)[/tex] has no real solutions. Therefore, the line and the circle do not intersect.
### Part (b): Find the coordinates of [tex]\( P \)[/tex] and the exact value of the shortest distance from [tex]\( P \)[/tex] to the circle.
To find the point [tex]\( P \)[/tex] on the line [tex]\( y = x - 5 \)[/tex] that is closest to the center of the circle (3, 4), we minimize the distance between a point on the line and the center of the circle.
Consider a point [tex]\( P \)[/tex] on the line given by the coordinates [tex]\( (x, y) \)[/tex] where [tex]\( y = x - 5 \)[/tex]. Let [tex]\( P = (x, x - 5) \)[/tex].
The distance [tex]\( D \)[/tex] between the point [tex]\( P \)[/tex] and the center of the circle [tex]\( (3, 4) \)[/tex] is given by the distance formula:
[tex]\[ D = \sqrt{(x - 3)^2 + ((x - 5) - 4)^2} \][/tex]
Simplify the expression inside the square root:
[tex]\[ D = \sqrt{(x - 3)^2 + (x - 9)^2} \][/tex]
To minimize this distance, we can consider the derivative. However, we can also directly use symmetry and geometric considerations. Since the equation of the line is linear and we're calculating a distance to a fixed point, the closest [tex]\( x \)[/tex] will be the perpendicular projection.
Minimize the distance, recognize that it happens when the derivative of [tex]\( D \)[/tex] with respect to [tex]\( x \)[/tex] is zero, find the critical points.
Solving the differentiation results:
[tex]\[ P = (6, 1) \][/tex]
Finally, calculate the exact distance from [tex]\( P = (6, 1) \)[/tex] to the center of the circle [tex]\( (3, 4) \)[/tex]:
[tex]\[ D = \sqrt{(6 - 3)^2 + (1 - 4)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \][/tex]
### Summary of Results:
(a) The line does not intersect the circle.
(b) The coordinates of [tex]\( P \)[/tex] are [tex]\( (6, 1) \)[/tex] and the exact value of the shortest distance from [tex]\( P \)[/tex] to the circle is [tex]\( 3\sqrt{2} \)[/tex].
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