Question 756440
{{{drawing(300,300,-6,2,-1,7,
grid(1),
blue(circle(-2,3,2)),
blue(circle(-2,3,0.1)),
green(line(-6,3,2,3)), locate(-5.9,3,green(y=3)),
locate(-1.9,3,O),locate(0.1,3,P)
)}}} The tangent to a circle is perpendicular to the radius at the point of tangency.
Since the y-axis is vertical, at the point of tangency, rhe radius of a circle tangent to the y-axis must be horizontal.
The horizontal line that passes through point O(-2,3) is {{{y=3}}}, so
1) a radius of the circle with center at (-2,3) and tangent to the y-axis is contained in {{{y=3}}}, and
2) the point of tangency of the circle with center at (-2,3) and the y-axis is the intersection of {{{y=3}}} and the y-axis.
That point of tangency is P(0,3).
The radius at that point is the segment OP, and its length is {{{0-(-2)=2}}}.
The equation of the circle with center at (-2,3) and radius 2 is
{{{(x+2)^2+(y-3)^2=2^2}}} --> {{{highlight((x+2)^2+(y-3)^2=4)}}}
because the equation of a circle with center at (h,k) and radius r is
{{{(x-h)^2+(y-k)^2=r^2}}},
but you do not need to remember that formula because
1) you may know that the distance from a point (x,y) to point O(-2,3) is
{{{sqrt((x+2)^2+(y-3)^2)}}} or
2) you may apply the Pytagorean theorem to the right triangle in the figure below to find that {{{(x+2)^2+(y-3)^2=2^2}}}
{{{drawing(300,300,-0.55,0.15,-0.05,0.65,
grid(0),
blue(circle(-0.2,0.3,0.2)),
blue(circle(-0.2,0.3,0.01)),
blue(circle(-0.1,0.473,0.01)),
blue(circle(-0.1,0.3,0.01)),
green(line(-6,0.3,2,0.3)),locate(-0.54,0.3,green(y=3)),
blue(triangle(-0.19,0.3,-0.1,0.473,-0.1,0.3)),
blue(rectangle(-0.1,0.3,-0.13,0.33)),
locate(-0.32,0.35,O(-2,3)),
locate(-0.15,0.53,R(x,y)),
locate(-0.13,0.3,S(x,3))
)}}} {{{system(OR^2=2^2,OS^2=(x+3)^2,SR^2=(y-3)^2)}}} --> {{{(x+2)^2+(y-3)^2=2^2}}}