The intersection point is only one. Then the equation of the line is tangent to the circle at point (1, -2).
<h3>What is a circle?</h3>
It is a locus of a point drawn an equidistant from the center. The distance from the center to the circumference is called the radius of the circle.
Prove algebraically that the straight line with equation x = 2y + 5 is a tangent to the circle with equation x² + y² = 5.
x = 2y + 5 ...1
x² + y² = 5 ...2
If the intersection of the point of the circle and line is one. Then the line is tangent to the circle.
Then from equations 1 and 2, we have
(2y + 5)² + y² = 5
4y² + 25 + 20y + y² - 5 = 0
5y² + 20y + 20 = 0
5y² + 10y + 10y + 20 = 0
5y (y + 2) + 10(y + 2) = 0
(5y + 10)(y + 2) = 0
y = -2, -2
Then the value of y is unique then the value of x will be unique.
The value of x will be
x = 2(-2) + 5
x = -4 + 5
x = 1
The intersection point is only one. Then the equation of the line is tangent to the circle at point (1, -2).
More about the circle link is given below.
brainly.com/question/11833983
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Answer:
The slope is 2.2
Step-by-step explanation:
Since the point goes through (0,0) the slope is given by
m =y/x
4.4 / 2
2.2
Answer:
It only has one solution
Step-by-step explanation:
Answer:
108°
Step-by-step explanation:
The reason for this is because opposite angles are equal
Yeah I think so I just took a wild guess and looked at my measurement sheet
