Check the picture below, so the hyperbola looks more or less like so, so let's find the length of the conjugate axis, or namely let's find the "b" component.
![\textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Ctextit%7Bhyperbolas%2C%20horizontal%20traverse%20axis%20%7D%20%5C%5C%5C%5C%20%5Ccfrac%7B%28x-%20h%29%5E2%7D%7B%20a%5E2%7D-%5Ccfrac%7B%28y-%20k%29%5E2%7D%7B%20b%5E2%7D%3D1%20%5Cqquad%20%5Cbegin%7Bcases%7D%20center%5C%20%28%20h%2C%20k%29%5C%5C%20vertices%5C%20%28%20h%5Cpm%20a%2C%20k%29%5C%5C%20c%3D%5Ctextit%7Bdistance%20from%7D%5C%5C%20%5Cqquad%20%5Ctextit%7Bcenter%20to%20foci%7D%5C%5C%20%5Cqquad%20%5Csqrt%7B%20a%20%5E2%20%2B%20b%20%5E2%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)

Answer:
No
Step-by-step explanation:
The equation of a circle with center (a,b) and radius r is given as:

If a given point (x,y) does not lie on this circle, it will not satisfy its equation.
This means the distance from the point to the center is not equal to the radius.
It is either less or greater than the radius.
Hence you cannot write the equation of the circle.
The solution to the given equation is x = ±7. The correct option is d. x = ±7
<h3>Solving quadratic equations </h3>
From the question, we are to determine the solution to the given equation
The given equation is
x²-49 = 0
NOTE: I assumed the variable is x
Solving the equation
x²-49 = 0
Writing in the form ax²= c
x² = 49
∴ x = ±√49
x = ±7
Hence, the solution to the given equation is x = ±7
Learn more on Solving quadratic equations here: brainly.com/question/24334139
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Answer: y=7
Step-by-step explanation: