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Find the equation of a circle, the end points of one of whose diameters are A (5, - 3) and B (2, - 4) ?

Find the equation of a circle, the end points of one of whose diameters are A (5, - 3) and B (2, - 4) ? Correct Answer x<span style="position: relative; line-height: 0; vertical-align: baseline; top: -0.5em;">2</span> + y<span style="position: relative; line-height: 0; vertical-align: baseline; top: -0.5em;">2</span> - 7x + 7y + 22 = 0

CONCEPT:

If (x1, y1) and (x2, y2) are the end points of the diameter of a circle. Then the equation of such a circle is (x – x1) ⋅ (x – x2) + (y – y1) (y – y2) = 0

CALCULATION:

Given: The end points of the diameter of a circle are A (5, - 3) and B (2, - 4).

As we know that, if (x1, y1) and (x2, y2) are the end points of the diameter of a circle then the equation of such a circle is (x – x1) ⋅ (x – x2) + (y – y1) (y – y2) = 0

Here, x1 = 5, y1 = - 3, x2 = 2 and y2 = - 4

⇒ (x - 5) ⋅ (x - 2) + (y + 3) ⋅ (y + 4) = 0

⇒ x2 + y2 - 7x + 7y + 22 = 0

So, the equation of the required circle is: x2 + y2 - 7x + 7y + 22 = 0

Hence, option B is the correct answer.

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