The equation of a circle that intersects the circle ${x^2} + {y^2} + 14x + 6y + 2 = 0$orthogonally and whose centre is $(0, 2)$ is
The equation of a circle that intersects the circle ${x^2} + {y^2} + 14x + 6y + 2 = 0$orthogonally and whose centre is $(0, 2)$ is
- A
${x^2} + {y^2} - 4y - 6 = 0$
- B
${x^2} + {y^2} + 4y - 14 = 0$
- C
${x^2} + {y^2} + 4y + 14 = 0$
- D
${x^2} + {y^2} - 4y - 14 = 0$
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