Answer:
(x-14)^2+(y-3)^2=9
Step-by-step explanation:
equation of a circle is (x-h)^2+(y-h)^2=r^2
so center is (14,3) and is tangent externally means
(x-14)^2+(y-3)2=3^2
(x-14)^2+(y-3)^2=9 answer
The equation of a circle is externally tangent to the given circle and has a center at (14, 3) is [tex](x-14)^2 + (y-3)^2=9[/tex]
The standard formula for finding the equation of a circle is expressed as:
[tex](x-a)^2+(y-b)^2=r^2[/tex]
where
(a, b) is the centre
r is the radius
Given the center at (14, 3)
If the equation of a circle is externally tangent to the given circle and has a center at (14, 3), then the radius will be 3
Substitute the radius and the centre into the expression above to have:
[tex](x-14)^2 + (y-3)^2=3^2\\(x-14)^2 + (y-3)^2=9[/tex]
Hence the equation of a circle is externally tangent to the given circle and has a center at (14, 3) is [tex](x-14)^2 + (y-3)^2=9[/tex]
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