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Tangents are drawn from (4, 4) to the circle x2 + y2 – 2x – 2y – 7 = 0 to meet the circle at A and B. The length of the chord AB is
  • a)
    2√3
  • b)
    3√2
  • c)
    2√6
  • d)
    6√2
Correct answer is option 'B'. Can you explain this answer?
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Tangents are drawn from (4, 4) to the circle x2+ y2–2x –2y...
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Tangents are drawn from (4, 4) to the circle x2+ y2–2x –2y...
= 25. Find the equations of the tangents and the points of contact.

We can begin by finding the center and radius of the circle. The equation x2 + y2 = 25 represents a circle with center (0, 0) and radius 5.

To find the tangent lines from (4, 4), we can use the fact that the tangent to a circle is perpendicular to the radius at the point of contact. So we need to find the radius that passes through (4, 4) and find its slope. The slope of the radius is given by (y2 – y1)/(x2 – x1) = (0 – 4)/(0 – 4) = 1.

Since the tangent is perpendicular to the radius, its slope is the negative reciprocal of the slope of the radius, which is -1. So we have the equation of the tangent line passing through (4, 4) as y – 4 = -1(x – 4), or y = -x + 8.

To find the point of contact, we need to find where this line intersects the circle. Substituting y = -x + 8 into the equation of the circle, we get:

x2 + (-x + 8)2 = 25
2x2 – 16x + 39 = 0

Solving for x using the quadratic formula, we get:

x = (16 ± √112)/2 = 8 ± 2√7

Substituting these values into the equation of the tangent line, we get the corresponding y-coordinates:

y = -x + 8 = 8 ± 2√7

So the two points of contact are (8 + 2√7, 8 – 2√7) and (8 – 2√7, 8 + 2√7).

Therefore, the equations of the tangents are y = -x + 8 and x + y = 16, and the points of contact are (8 + 2√7, 8 – 2√7) and (8 – 2√7, 8 + 2√7).
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Tangents are drawn from (4, 4) to the circle x2+ y2–2x –2y –7 = 0 to meet the circle at A and B. The length of the chord AB isa)2√3b)3√2c)2√6d)6√2Correct answer is option 'B'. Can you explain this answer?
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