Anonymous

(x-3)^2 + (y+5)^2 = 13^2 First find the slope. This is dy/dx You have 2(x-3) + 2 (y+5) dy/dx=0 dy/dx= -2(x-3) /2(y+5). At x=15 and y=0, we have dy/dx= -*12/5= -12/5. You have y = dy/dx x +c= -12/5 x +c . c is a constant. This is the equation of line.

Anonymous

C (3 , - 5) , P (15 , 0) m CP = 5 / 12 m tangent = - 12 / 5 y = (-12/5) ( x - 15 ) y = (-12/5) x + 36_______equation of tangent at P (15,0)

Anonymous

(x-3)² + (y+5)² = 13² center of circle is at C(3,-5) slope of CP = (-5-0)/(3-15) = 5/12 slope of tangent at P(15,0) = -12/5 = -2.4 point-slope equation of tangent at P(15,0): y-0 = -2.4(x-15) y = -2.4x + 36