Can someone explain to me on how to solve this problem? :)
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Distance from point (x₁, y₁) to line Ax + By + C = 0
d = |Ax₁ + By₁ + C| / √(A²+B²)
Radius = distance from centre point (h, k) to line x − 3y + 8 = 0
r = |h − 3k + 8| / √10
Radius = distance from centre point (h, k) to line 3x − y = 0
r = |3h − k| / √10
Radius = distance from centre point (h, k) to point (3,7)
r = √((h−3)² + (k−7)²)
So we get 3 equations:
(1) r = |h − 3k + 8| / √10
(2) r = |3h − k| / √10
(3) r = √((h−3)² + (k−7)²)
From (1) and (2) we get:
|h − 3k + 8| = |3h − k|
h − 3k + 8 = 3h − k .... or .... h − 3k + 8 = −(3h − k)
k = 4 − h .... or .... k = h + 2
——————————————————————————————
Case 1: k = 4−h
Substituting this value of k in equation (1) or (2) we get: r = |4h−4|/√10
Substituting this value of k in equation (3) we get: r = √(2h²+18)
|4h−4|/√10 = √(2h²+18)
(4h−4)²/10 = 2h² + 18
16h² − 32h + 16 = 20h² + 180
4h² + 32h + 164 = 0
4 (h² + 8h + 41) = 0
NO real solutions.
——————————————————————————————
Case 2: k = h+2
Substituting this value of k in equation (1) or (2) we get: r = |2h−2|/√10
Substituting this value of k in equation (3) we get: r = √(2h²−16h+34)
|2h−2|/√10 = √(2h²−16h+34)
(2h−2)²/10 = 2h² − 16h + 34
4h² − 8h + 4 = 20h² − 160h + 340
16h² − 152h + 336 = 0
8 (2h² − 19h + 42) = 0
8 (2h − 7) (h − 6) = 0
h = 7/2, h = 6
h = 7/2 ----> k = 7/2+2 = 11/2 ----> r = |2*7/2−2|/√10 = 5/√10 = √(5/2)
h = 6 ------> k = 6 + 2 = 8 --------> r = |2*6−2|/10 = 10/√10 = √10
Equations of circles:
(x − 7/2)² + (y − 11/2)² = 5/2
(x − 6)² + (y − 8)² = 10
d = |Ax₁ + By₁ + C| / √(A²+B²)
Radius = distance from centre point (h, k) to line x − 3y + 8 = 0
r = |h − 3k + 8| / √10
Radius = distance from centre point (h, k) to line 3x − y = 0
r = |3h − k| / √10
Radius = distance from centre point (h, k) to point (3,7)
r = √((h−3)² + (k−7)²)
So we get 3 equations:
(1) r = |h − 3k + 8| / √10
(2) r = |3h − k| / √10
(3) r = √((h−3)² + (k−7)²)
From (1) and (2) we get:
|h − 3k + 8| = |3h − k|
h − 3k + 8 = 3h − k .... or .... h − 3k + 8 = −(3h − k)
k = 4 − h .... or .... k = h + 2
——————————————————————————————
Case 1: k = 4−h
Substituting this value of k in equation (1) or (2) we get: r = |4h−4|/√10
Substituting this value of k in equation (3) we get: r = √(2h²+18)
|4h−4|/√10 = √(2h²+18)
(4h−4)²/10 = 2h² + 18
16h² − 32h + 16 = 20h² + 180
4h² + 32h + 164 = 0
4 (h² + 8h + 41) = 0
NO real solutions.
——————————————————————————————
Case 2: k = h+2
Substituting this value of k in equation (1) or (2) we get: r = |2h−2|/√10
Substituting this value of k in equation (3) we get: r = √(2h²−16h+34)
|2h−2|/√10 = √(2h²−16h+34)
(2h−2)²/10 = 2h² − 16h + 34
4h² − 8h + 4 = 20h² − 160h + 340
16h² − 152h + 336 = 0
8 (2h² − 19h + 42) = 0
8 (2h − 7) (h − 6) = 0
h = 7/2, h = 6
h = 7/2 ----> k = 7/2+2 = 11/2 ----> r = |2*7/2−2|/√10 = 5/√10 = √(5/2)
h = 6 ------> k = 6 + 2 = 8 --------> r = |2*6−2|/10 = 10/√10 = √10
Equations of circles:
(x − 7/2)² + (y − 11/2)² = 5/2
(x − 6)² + (y − 8)² = 10