PLEASE help really quick! (; Math problem.
1 answer:
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Answer:
a). Center of the circle = (-2, 5)
b). Equation of the line ⇒ y =
Step-by-step explanation:
Equation of the circle is,
x² + 4x + y²- 10y + 20 = 30
a). [x² + 2(2)x + 4 - 4] + [y²- 2(5)y + 25] - 25 + 20 = 30
[x² + 2(2)x + 4] - 4 + [y² - 2(5)y + 25] - 25 + 20 = 30
(x + 2)² + (y - 5)²- 29 + 20 = 30
(x + 2)² + (y - 5)²- 9 = 30
(x + 2)² + (y - 5)² = 39
By comparing this equation with the standard equation of a circle,
Center of the circle is (-2, 5).
b). A point (2, 10) lies on this circle.
Slope of the line joining this point to the center (-2, 5),
=
=
Let the slope of the tangent which is perpendicular to this line is ''
Then by the property of perpendicular lines,
Now the equation of the line passing though (2, 10) having slope
y - y' =
y - 10 =
y - 10 =
y =
y =
Therefore, equation of the line will be, y =
Answer:
y=-3x+5
Step-by-step explanation:
concepts: y=mx+b is slope intercept equation formula
m=slope
b= y intercept
Therefore we need to find the slope and y intercept
first find slope
m=
let y2 be 2
let y1 be -4
let x2 be 1
let x1 be 3
6/-2 = -3
m=-3
slope is -3
y= -3x+ b
we need find y intercept now
just plug in (1,2) into that equation
2=-3(1)+b
b=5
y=-3x+5