Question

find the circle and radius of the circle x^2+y^2+10x+8y+5=0​

Answer 1

Answer:

There are 2 tangents with slope 1

There are 2 tangents with slope 1   x-y + 6√2 - 1 = 0 and

There are 2 tangents with slope 1   x-y + 6√2 - 1 = 0 and   x-y - 6√2 - 1 = 0

There are 2 tangents with slope 1   x-y + 6√2 - 1 = 0 and   x-y - 6√2 - 1 = 0Step-by-step explanation:

There are 2 tangents with slope 1   x-y + 6√2 - 1 = 0 and   x-y - 6√2 - 1 = 0Step-by-step explanation:Equation of line with given slope m , and y-intercept c is given by

There are 2 tangents with slope 1   x-y + 6√2 - 1 = 0 and   x-y - 6√2 - 1 = 0Step-by-step explanation:Equation of line with given slope m , and y-intercept c is given byy = mx+c.

There are 2 tangents with slope 1   x-y + 6√2 - 1 = 0 and   x-y - 6√2 - 1 = 0Step-by-step explanation:Equation of line with given slope m , and y-intercept c is given byy = mx+c.Given m =1, thus tangent would be in the form

There are 2 tangents with slope 1   x-y + 6√2 - 1 = 0 and   x-y - 6√2 - 1 = 0Step-by-step explanation:Equation of line with given slope m , and y-intercept c is given byy = mx+c.Given m =1, thus tangent would be in the formy = x +c

There are 2 tangents with slope 1   x-y + 6√2 - 1 = 0 and   x-y - 6√2 - 1 = 0Step-by-step explanation:Equation of line with given slope m , and y-intercept c is given byy = mx+c.Given m =1, thus tangent would be in the formy = x +cFor the given circle , centre is (5, 4) and radius is 6.

There are 2 tangents with slope 1   x-y + 6√2 - 1 = 0 and   x-y - 6√2 - 1 = 0Step-by-step explanation:Equation of line with given slope m , and y-intercept c is given byy = mx+c.Given m =1, thus tangent would be in the formy = x +cFor the given circle , centre is (5, 4) and radius is 6.For any given line to be a tangent to any given circle, the perpendicular distance from the centre of the circle to tangent line should be equal to radius of the circle.

There are 2 tangents with slope 1   x-y + 6√2 - 1 = 0 and   x-y - 6√2 - 1 = 0Step-by-step explanation:Equation of line with given slope m , and y-intercept c is given byy = mx+c.Given m =1, thus tangent would be in the formy = x +cFor the given circle , centre is (5, 4) and radius is 6.For any given line to be a tangent to any given circle, the perpendicular distance from the centre of the circle to tangent line should be equal to radius of the circle.Thus, |5-4 +c| /√2 = 6, On cross multiplication, we get

There are 2 tangents with slope 1   x-y + 6√2 - 1 = 0 and   x-y - 6√2 - 1 = 0Step-by-step explanation:Equation of line with given slope m , and y-intercept c is given byy = mx+c.Given m =1, thus tangent would be in the formy = x +cFor the given circle , centre is (5, 4) and radius is 6.For any given line to be a tangent to any given circle, the perpendicular distance from the centre of the circle to tangent line should be equal to radius of the circle.Thus, |5-4 +c| /√2 = 6, On cross multiplication, we get|c + 1| = 6√2,

There are 2 tangents with slope 1   x-y + 6√2 - 1 = 0 and   x-y - 6√2 - 1 = 0Step-by-step explanation:Equation of line with given slope m , and y-intercept c is given byy = mx+c.Given m =1, thus tangent would be in the formy = x +cFor the given circle , centre is (5, 4) and radius is 6.For any given line to be a tangent to any given circle, the perpendicular distance from the centre of the circle to tangent line should be equal to radius of the circle.Thus, |5-4 +c| /√2 = 6, On cross multiplication, we get|c + 1| = 6√2,c + 1 = ±6√2,

There are 2 tangents with slope 1   x-y + 6√2 - 1 = 0 and   x-y - 6√2 - 1 = 0Step-by-step explanation:Equation of line with given slope m , and y-intercept c is given byy = mx+c.Given m =1, thus tangent would be in the formy = x +cFor the given circle , centre is (5, 4) and radius is 6.For any given line to be a tangent to any given circle, the perpendicular distance from the centre of the circle to tangent line should be equal to radius of the circle.Thus, |5-4 +c| /√2 = 6, On cross multiplication, we get|c + 1| = 6√2,c + 1 = ±6√2,c = 6√2-1 or -6√2-