Question

equation of tangent having slope 1 to circle x^2+y^2-10x-8y+5=0

Answer 1

Answer:

Step-by-step explanation:

Given equation of the circle x^2+y^2-10x-8y+5=0

Center of the circle is ( 5, 4)

Radius of the given circle is 6

Now,

let the tangent be y = mx + c

where m slope of the tangent and

c the y-intercept.

Now, given the slope of the tangent is 1

=> m =1,

Hence, tangent is y = x + c.

We know that perpendicular distance from the center of the circle onto the

tangent to the circle will be equal to the radius of the circle.

=> |5 - 4 + c|/√2 = 6

=> | 1 + c| = 6√2

=> c + 1 = ± 6√2

=> c  = ±6√2 - 1.

Hence there are 2 tangents with slope 1.

y = x + 6√2 - 1

and

y = x - 6√2 - 1.