If the length of the tangent from the point (a,1) to the circle x²+y²+5x+7y+3=0 is 5, find the value of a.?

The circle x²+y²+5x+7y+3=0 is centred at (-5/2, -7/2) and has the radius whose square is (¼a²+¼b²-c) which is ¼(5²+7²-4×3) = 31/2

The square of 5 is 25, so the square of the distance from the external tangent point to the centre of the circle is 25 + 31/2 = 81/2.

(a+5/2)² + (1+7/2)² = 81/2 → (a+5/2)² - 81/4 = 0 → (a+5/2+9/2)(a+5/2-9/2) = 0 → (a+7)(a-2) = 0

Solutions: a = -7 and a = 2

Find the radius of the circle

x^2 + 5x + y^2 + 7y = -3

x^2 + 2 * (5/2) * x + y^2 + 2 * (7/2) * y = -3

x^2 + 2 * (5/2) * x + (5/2)^2 + y^2 + 2 * (7/2) * y + (7/2)^2 = -3 + (5/2)^2 + (7/2)^2

(x + (5/2))^2 + (y + (7/2))^2 = -3 + 25/4 + 49/4

(x + (5/2))^2 + (y + (7/2))^2 = 74/4 - 12/4

(x + (5/2))^2 + (y + (7/2))^2 = 62/4

(x + (5/2))^2 + (y + (7/2))^2 = sqrt(62) / 2

The radius of the circle is sqrt(62) / 2

The distance from (a , 1) to the tangent point is 5

The line from the center of the circle to the tangent point meets at a right angle to the line from (a , 1) to the tangent point

(sqrt(62)/2)^2 + 5^2 = p^2

62/4 + 25 = p^2

62/4 + 100/4 = p^2

162/4 = p^2

(a - (5/2))^2 + (1 - (7/2))^2 = 162/4

(a - (5/2))^2 + (-5/2)^2 = 162/4

(a - (5/2))^2 + 25/4 = 162/4

(a - (5/2))^2 = 137/4

a - 5/2 = +/- sqrt(137) / 2

a = (5 +/- sqrt(137)) / 2