Respuesta :

Answer:

[tex]\implies x^2+y^2-4x -6y+9=0[/tex]

Step-by-step explanation:

Circle is touching y-axis at point (0, 3) and is centered at (2, 3)

-> h = 2, k = 3 & r = 2 units (If a circle touches y -axis then x coordinate of the center will be it's radius)

Equation of circle in standard form is given as:

[tex](x-h)^2 +(y-k)^2 =r^2[/tex]

Plugging the values of h, k and r in the above equation, we find:

[tex](x-2)^2 +(y-3)^2 =2^2[/tex]

[tex]\implies x^2-4x +4+y^2-6y+9=4[/tex]

[tex]\implies x^2+y^2-4x -6y+4+9=4[/tex]

[tex]\implies x^2+y^2-4x -6y+9=0[/tex]

This is the required equation of the circle.

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