Please help me to solve this equation

Question

Write an equation in slope intercept from that is tangent to the circle (x-2)²+(y-3)²=5 and has negative slope where the circle meets the y axis.

Answer 1

The circle intersects the y-axis at the point (0,y).

So, (0-2)² + (y-3)² = 5

Solving for y gives y = 2 or 4

The points of tangency are (0,4) and (0,2).

At (0,2) the tangent line has negative slope and is perpendicular to the line through the center (2,3).

Slope of line through (2,3) and (0,2) is (2-3) / (0-2) = 1/2

So, slope of tangent line is -1 / (1/2) = -2

Equation of tangent line: y - 2 = -2( x - 0)

y = -2x + 2

Answer 2

the circle meets the y axis when x=0

(-2)^2 +(y-3)^2 = 5

(y-3)^2 = 5-4 =1

y-3 = + or -1

y = 3+1 or 3-1 = 4 or 2

the points are (0,2) and (0,4)

the circle (x-2)^2 + (y-3)^2 = 5

is x^2 -4x +4 + y^2 -6x +9 =5

or x^2 -4x +y^2 -6y = 5-9-4 =-8

take the derivative

2x -4 +2yy' -6y' =0

solve for y'

y' = (4-2x)/(2y-6)

at (0,2) y' = 4/-2=-2 = slope

tangent line is y=-2x+2 in slope intercept form

at (0,4) y' = 4/2=2

tangent line is y=2x + 4

two tangent lines

y = 2x+4 at (0,4) and y=-2x+4 at (0,2)

it helps to sketch the circle and see where it intersects the y axis, at two points

just a rough sketch shows one tangent line is downward sloping and the other is upward sloping

the tangent line with negative slope is at the circle point (0,2): y=-2x+4

that's on the lower left side of the circle