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[[1]] is the equation of the circle concentric with the circle $x^{2} + y^{2} - 8 x + 6 y - 5 = 0$ and passing through the point (−2,−7).

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Detailed Solution

We are given an equation of a circle

x^{2} + y^{2} - 8 x + 6 y - 5 = 0

We have to find the equation of another circle which is concentric with the above circle and passes through (−2,−7)
seo

Concentric circles have the same centre so find the centre of the given circle.
On comparing

x^{2} + y^{2} - 8 x + 6 y - 5 = 0

with

x^{2} + y^{2} + 2 gx + 2 fy + c = 0

, we get the value of g is -4 and value of f is 3.
Centre is (−g,−f)=(−(−4),−3)=(4,−3)
seo

The radius of the required circle will be the distance between centre and (−2,−7).
Radius is

r = \sqrt{{(4 - (- 2))}^{2} + {(- 3 - (- 7))}^{2}}

\to r = \sqrt{6^{2} + 4^{2}} = \sqrt{36 + 16}

\to r = \sqrt{52} units

The equations of the new circle is

{(x - h)}^{2} + {(y - k)}^{2} = r^{2}

, where r is the radius,
(h, k) are the coordinates of centre.

{(x - h)}^{2} + {(y - k)}^{2} = r^{2}

r = \sqrt{52}, (h, k) = (4, - 3)

\to {(x - 4)}^{2} + {(y - (- 3))}^{2} = {(\sqrt{52})}^{2}

\to {(x - 4)}^{2} + {(y + 3)}^{2} = 52

\to x^{2} + 16 - 8 x + y^{2} + 9 + 6 y = 52

\to x^{2} + y^{2} - 8 x + 6 y + 25 = 52

∴ x^{2} + y^{2} - 8 x + 6 y - 27 = 0

Therefore, the equation of the circle is

x^{2} + y^{2} - 8 x + 6 y - 27 = 0

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