How do you write the standard equation of the circle the given center that passes with through the given point: center (5,9); point (2, 9)?

Answer 1

#(x-5)^2+(y-9)^2=3^2, or, x^2+y^2-10x-18y-97=0.#

The Centre #C(h,k)# of the Circle is #C(h,k)=C(5,9)# and the
circle passes through a point #P(2,9).#

Hence, using the Distance Formula, the Radius

#r=CP=sqrt{(5-2)^2+(9-9)^2}=3.#
We know that, the Eqn. of the circle having centre at #(h,k)# and
radius #r# is, #(x-h)^2+(y-k)^2=r^2.#

Therefore, the Reqd. Eqn. is,

#(x-5)^2+(y-9)^2=3^2, or, x^2+y^2-10x-18y-97=0.#
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Answer 2

The standard equation of the circle with center (5, 9) that passes through the point (2, 9) is:

(x - 5)^2 + (y - 9)^2 = 9

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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