Question

find the equation of circle passing through the poi (-4,2)and touching lines x+y=2 and x-y=2

Answer 1

Let equation of circle is x² + y² + 2gx + 2fy + c = 0 so, centre = (-g, -f) and radius =√{g² + f² - c}

now, circle passing through (-4,2)
so, (-4)² + (2)² + 2g(-4) + 2f(2) + c = 0
20 - 8g + 4f + c = 0 --------(1)

circle touch the lines x + y = 2 and x - y = 2
so, use concept P = r , where P is the distance between line and centre .
|-g -f- 2|/√2 = √{g² + f² - c} ---------(2)
take square both sides,
(g + f + 2)² = 2(g² + f² - c) --------(3)

similarly ,
|-g+f-2|/√2=(g² + f² - c)------------(4)
(f - g - 2)² = 2(g² + f² - c)²--------(5)
from equations (3) and (5)

(g + f + 2)² = (f - g - 2)²
(g + f + 2 - f + g + 2)(g + f + 2 + f - g - 2)= 0
(4 + 2g)(f) = 0
f = 0, g = -2


now, we have 2 cases .
case 1 :- when f = 0
from equations (2) ,
| -g - 2|/√2 =√(g² - c) => (g + 2)² = 2(g² - c)
g² + 4g + 4 = 2g² - 2c
g² -4g -2c - 4 = 0 ---------(6)

now put f = 0 in equation (1)
20 - 8g + 4 × 0 + c = 0
20 - 8g + c = 0 ----------(7)

now, solve equations (6) and (7)
g²- 4g - 2(8g - 20) - 4 = 0
g² - 4g -16g + 36 = 0
g² - 20g + 36 = 0 => g = 2, 18
and c = 8g - 20 = -4, 124

then, equation of circles are
x² + y² + 2×2x + 2×0 + (-4) = 0
x² + y² + 4x - 4 = 0

and x² + y² + 2×18x + 0 + (124) = 0
x² + y² + 36x + 124 = 0


case 2 :- when g = -2
from equation (2)
f² = 2(4 + f² - c )
f² -2c + 8 = 0 -------(8)
now, put g = -2 in equation (1)
20 - 8(-2) + 4f + c = 0
20 + 16 + 4f + c = 0
4f + c + 36 = 0 --------(9)
solve equations (8) and (9)
f² - 2(-36-4f) + 8 = 0
f² + 72 + 8f + 8 = 0 => f² + 8f + 80 = 0
f = {-8 ± √(64 - 320)}/2 = -4 ± 8i { complex number }
c = -36 - 4(-4±8i)
$= -36 +16 \mp32i$

$= -20\mp32i$


now, equation of circles are
${x}^{2} + {y}^{2} + 2(-2)x + 2(-4\pm8i)y+(-20\mp32i)$