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प्रश्न
Find the equation of a circle which touches x-axis at a distance 5 from the origin and radius 6 units.
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उत्तर
The given circle touches the x-axis at the point (5, 0).
So, it lies in five quadrant.
When a circle touches x-axis, then its radius is equal to the absolute value of the y-coordinates of the centre.
So the centre of the circle is of the form (a, 6).
We know that the radius of the circle is the distance between the center and any point on its boundary.
So, by distance formula - (a - 5)2 + (6 - 0)2 = 62
⇒ a = 5 ...(Circle lies in first quadrant)
Hence, the centre of the circle is (5, 6).
The standard equation of the circle with centre (p, q) and the radius r is given by (x - p)2 + (y - q)2 = r2
Thus, the required circle equation is given as (x - 5)2 + (y - 6)2 = 62 or x2 + y2 - 10x - 12y + 25 = 0
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