Finding the equation......?

Question

Find the equation of the circle if the circle is tangent to the line x+2y=3 at the point (-1,2) and the center is on the y-axis.

Answer 1

The equation is x^2 + (y - 4)^2 = 5.
You can obtain the result if you take the standard equation of a circle in the form x^2 + (y - h)^2 = R^2 /center - (0, h) on the Y-axis, radius R/.
Now the system of both equations:
x + 2y = 3; x^2 + (y - h)^2 = R^2 must have exactly 1 solution for the line given to be tangent, and, taking into account the touching point (-1,2) we have the relationship
R^2 = 1 + (h - 2)^2 from the corresponding right triangle.
Putting x = 3 - 2y in the second equation above we obtain:
(3 - 2y)^2 + (y - h)^2 = 1 + (h- 2)^2,
the latter having exactly one solution when its discriminant
4(h + 6)^2 - 4*5(4 + 4h) = 0, leading to
h^2 - 8 h +16 = 0, h = 4 and then R^2 = 5

P.S. The coordinates (-1, 2) do not satisfy the equation
x^2 + y^2 - 7y = -10 in the next answer, so it is not correct.

P.S.(2) Now it is corrected.

Answer 2

first we need to find the equation of the line that is perpendicular to the given line and passes through the given point.

x+2y=3

2y=-x+3

y=(-1/2)x+3/2

the original slope is -1/2
the perpendicular slope is the opposite reciprocal of the original or 2.

y=2x+b...plug in the given point for x and y.

2=2(-1)+b...solve for b.

b=4...this makes (0,4) the y intercept and since the center of the circle lies on the y axis it is also your center.

now plug the center and given point into the equation of a circle...(x-h)^2+(y-k)^2=r^2 and simplify the left side.

(-1-0)^2+(2-4)^2=r^2
=>1+4=r^2
=>5=r^2...you don't need to solve for r unless you are graphing.

the equation is obtained by plugging h, k, and r^2 into the circle equation and simplifying.

(x-0)^2+(y-4)^2=5
or
x^2+(y-4)^2=5

Answer 3

2y = - x + 3
y = (- 1 / 2) x + 3 / 2
m1 = (-1/2)
m2 is gradient of radius that passes thro`(-1,2)
m1 x m2 = - 1
(-1/2) x m2 = - 1
m2 = 2
Gradient of radius at (-1 , 2) = 2
Radius passes thro` (0 , b) and (-1 , 2) and has gradient (slope) 2
2 = (2 - b) / (-1 - 0)
- 2 = 2 - b
b = 4
Centre (0 , 4)
P (-1 , 2) is on circumference
r² = 1² + 2²
r² = 5
Equation of circle is:-
x² + (y - 4)² = 5

Answer 4

y= -x/2+3/2

2

Answer 5

Center is (0,4)

Radius is 2.236

circle crosses the y intercept at (0, 6.236) an (0, 1.7634)

(x-0)^2 + (y-4)^2 = (2.236) ^2

x^2 + y^2 - 8y +16 = 5

x^2 + y^2 -8y = -11

Answer 6

x