# What is the equation of circles touching x-axis at the origin and the line 4x-3y+24=0

Aman Bansal
592 Points
11 years ago

Dear Diwakar,

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

Cracking IIT just got more exciting,It s not just all about getting assistance from IITians, alongside Target Achievement and Rewards play an important role. ASKIITIANS has it all for you, wherein you get assistance only from IITians for your preparation and win by answering queries in the discussion forums.

So start the brain storming…. become a leader with Elite Expert League ASKIITIANS

Thanks

Aman Bansal

Suzuki Mitsuba
4 Points
11 years ago

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)plz give me the expln of red line