Equation of the straight line meeting the circle with centre at origin and radius equal to 5 in two points at equal distances of 3 units from the point A(3, 4), isa)6x + 8y – 41 = 0b)6x – 8y + 41 = 0c)8x + 6y + 41 = 0d)8x – 6y + 41 = 0Correct answer is option 'A'. Can you explain this answer?

Question

Equation of the straight line meeting the circle with centre at origin and radius equal to 5 in two points at equal distances of 3 units from the point A(3, 4), isa)6x + 8y &ndash; 41 = 0b)6x &ndash; 8y + 41 = 0c)8x + 6y + 41 = 0d)8x &ndash; 6y + 41 = 0Correct answer is option 'A'. Can you explain this answer?

Answer

The equation of the straight line can be found by using the formula for the distance between a point and a line.

Let the equation of the straight line be y = mx + c.

The distance between the point A(3, 4) and any point on the line is given by:
d = |(mx + c) - 4| / sqrt(1 + m^2)

Since the line intersects the circle at two points at equal distances of 3 units from A(3, 4), we have:
|(mx + c) - 4| / sqrt(1 + m^2) = 3

Squaring both sides of the equation, we get:
(mx + c - 4)^2 / (1 + m^2) = 9

Expanding and simplifying the equation, we have:
(m^2 + 1)x^2 + 2mcx + (c^2 - 8c + 7) = 0

Since the line intersects the circle, the equation of the circle is:
x^2 + y^2 = 5^2
x^2 + y^2 = 25

Substituting y = mx + c into the equation of the circle, we get:
x^2 + (mx + c)^2 = 25

Expanding and simplifying the equation, we have:
x^2(1 + m^2) + 2mcx + c^2 = 25

Comparing the coefficients of the x^2 terms in the equations of the line and circle, we get:
1 + m^2 = 1 + m^2

Comparing the coefficients of the x terms in the equations of the line and circle, we get:
2mc = 2mc

Comparing the constant terms in the equations of the line and circle, we get:
c^2 - 8c + 7 = 25

Simplifying the constant term equation, we get:
c^2 - 8c - 18 = 0

Factoring the constant term equation, we get:
(c - 9)(c + 2) = 0

Solving for c, we get c = 9 or c = -2.

Therefore, the equation of the straight line can be:
y = mx + 9 or y = mx - 2.

So, the equation of the straight line meeting the circle with centre at origin and radius equal to 5 in two points at equal distances of 3 units from the point A(3, 4) is:
y = mx + 9 or y = mx - 2.

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