A circle having centre in the first quadrant and radius 3 units, touches the yy-axis at the point (0,3) .
This circle has equation
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Given,
The radius of the circle = 3 units
Centre = First quadrant
Y-intercept = 3
To find,
The equation of the circle.
Solution,
The equation of the circle is x² +y²-6y.
We can simply find the equation of the circle using the given condition.
(x-h)²+(y-k)² = r²
(x-0)² +(y-3)² = 9
x² +y² +9 - 6y = 9
x² +y²-6y
which is the required equation of the circle.
Hence, x² +y²-6y is the required equation of the circle whose center lies in the first quadrant and radius 3 units, touches the y-axis at the point (0,3).
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