Find the tangent equation passing through point a (- 4,3) and tangent to circle (x + 1) square + (y + 1) square = 25 Find the tangent equation passing through point a (- 4,3) and tangent to circle (x + 1) square + (y + 1) square = 25

Find the tangent equation passing through point a (- 4,3) and tangent to circle (x + 1) square + (y + 1) square = 25 Find the tangent equation passing through point a (- 4,3) and tangent to circle (x + 1) square + (y + 1) square = 25

Point a (- 4,3) is on a known circle, because the slope of the line between a and the center of the circle k = [3 - (- 1)] / [- 4 - (- 1)] = 4 / - 3 = - 4 / 3
So the tangent slope k = 3 / 4, the tangent equation is Y-3 = 3 / 4 (x + 4), that is Y-3 = 3x / 4 + 3, that is y = 3x / 4 + 6

Find the tangent equation through point a (3,4) and tangent to circle x square + y square = 25

Point a (- 4,3) is on a known circle, because the slope of the line between a and the center of the circle k = [3 - (- 1)] / [- 4 - (- 1)] = 4 / - 3 = - 4 / 3
So the tangent slope k = 3 / 4, the tangent equation is Y-3 = 3 / 4 (x + 4), that is Y-3 = 3x / 4 + 3, that is y = 3x / 4 + 6
I hope it can help you

Tangent equation passing through point (1, - 7) and tangent to circle x2 + y2 = 25______ ．

If the slope of the tangent does not exist, because the tangent passes through the point (1, - 7), the linear equation is x = 1 and the circle x2 + y2 = 25 & nbsp; If the slope of the tangent exists, let the slope of the tangent be K. since the tangent passes through the point (1, - 7), let the equation of the tangent be y + 7 = K (x + 1), that is, kx-y + k-7 = 0. From the tangent line to the circle, the distance d from the center of the circle to the straight line is equal to the radius r, and the solution is k = - 34, or K = 43, so the equation of the tangent is 3x + 4Y + 25 = 0 or 4x-3y-25 = 0, so the answer is 3x + 4Y + 25 = 0 or 4x-3y-25 = 0