# Equation of circles touching x-axis at the origin and the line 4x – 3y + 24 = 0 are

Arun
25750 Points
6 years ago
Dear Student

Let the circle has centre (h, k) and radius r, so the eqn of the circle would be
(x - h)² + (y - k)² = r²
SInce points ((1, -1) and (37/13, 11/13) lies on the circle , they should satisfy the eqn of circle.
(1 - h)² + (-1 - k)² = r² ---------------1) , and for other points
(37/13 - h)² + (11/13 - k)² = r² -----2)

equate the radius (r²) from eqn 1) and 2)
(1 - h)² + (-1 - k)² = (37/13 - h)² + (11/13 - k)² ; or
1 - 2h + h² + 1 + 2k + k² = (37/13)² - 2*(37/13)*h + h² + (11/13)² - 2*(11/13)*k + k²
1 - 2h + 1 + 2k = (37/13)² - 2*(37/13)*h + (11/13)² - 2*(11/13)*k
74h/13 - 2h + 22k/13 + 2k = (37/13)² + (11/13)² - 2

48h/13 + 48k/13 = 1152/169 ; or
h + k = 1152*13)/(48*169) or 24/13 -----------3)

Also we know distance from tangent of a circle to its centre is its radius

distance = |ah + bk + c| / √(a² + b²)
r = | h + k - 48/13 | / √(1 + 1) ; or
r² = [(h + k) - 48/13]² / (√2)², plug h + k = 24/13 from eqn 3)
r² = [24/13 - 48/13]² / 2 or 288/169
r = 12√2 / 13

Now find distance from any point say (37/13, 11/13) to the centre (h, k) which is r
(11/13 - k)² + (37/13 - h)² = 288/169
(11 - 13k)² + (37 - 13h)² = 288 ; cancel out 169 at Denominator, and plug 13k = 24 - 13h ; eqn 3)
(11 - 24 +13h)² + (37 - 13h)² = 288
(13h - 13)² + (37 - 13h)² = 288
169h² - 338h + 169 + 37² - 962h + 169h² = 288
338h² - 1300h + 1250 = 0
h = [1300 ± √1690000 - 1690000] / 2*338 = 100/2*26 or 25/13
since h + k = 24/13, therefore k = 24/13 - 25/13 = -1/13
so centre of circle (h, k) would be )25/13, -1/13) and radius of circle would be r = 12√2/13
and eqn. of the circle would be

(x - h)² + (y - k)² = r² ; or
(x - 25/13)² + (y + 1/13)² = 288/139 ; or
(13x - 25)² + (13y + 1)² = (12√2)²

Regards