(x−2)² + (y+3)² = 25
x-int = (4, 0), y-int = (0, 3)
(−1−5)/(4−1) = −6/3 = −2
Tangent eqn: 3x+4y = 25
Substitute y = 2x−1 ⇒ x²+(2x−1)²=10 ⇒ x=1, −1 ⇒ (1, 1), (−1, −3)
Perpendicular distance = |0−4|/√2 = 2√2 ≠ √8 ⇒ Not tangent
y+2 = 2(x−1) ⇒ y = 2x − 4
Same slope ⇒ y+3 = (2/5)(x−4)
√((8−2)²+(9−3)²) = √72 = 6√2
(3²+4²)=25 ⇒ on the circle
Using slope equality ⇒ (k−2)/(3−1) = (6−k)/(5−3) ⇒ k = 4
Center = (2, −3), r = √(4+9+3) = √16 = 4
Midpoint = ((−4+10)/2, (7+−5)/2) = (3, 1)
Center = (2, 6), r = √((6−(−2))²+(8−4)²)/2 = √80/2 = 2√5 ⇒ (x−2)² + (y−6)² = 20
y−3 = 2(x−2) ⇒ y = 2x − 1