Question

Question 13 Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.

Class X1 - Maths -Conic Sections Page 241

Answer 1

The circle intercept the co-ordinate axes at a and b. it means x - intercept at ( a, 0) and y-intercept at (0, b) .

now, we observed that circle passes through points (0, 0) , (a, 0) and (0, b) .

we also know,
General equation of circle is
x² + y² + 2gx + 2fy + C = 0

when point (0,0)
(0)² + (0)² + 2g(0) + 2f(0) + C = 0
0 + 0 + 0 + 0 + C = 0
C = 0 -------(1)

when point (a,0)
(a)² + (0)² + 2g(a) + 2f(0) + C = 0
a² + 2ag + C = 0
from equation (1)
a² + 2ag = 0
a(a + 2g) = 0
g = -a/2

when point ( 0, b)
(0)² + (b)² + 2g(0) + 2f(b) + C = 0
b² + 2fb + C = 0
f = -b/2

now,
equation of circle is
x² + y² + 2x(-a/2) + 2y(-b/2) + 0 = 0 { after putting values of g, f and C }
x² + y² -ax -by = 0

Answer 2

Answer:

Step-by-step explanation:

Let the equation of the required circle be (x−h)2+(y−k)2=r2

Since the centre of the circle passes through (0,0)

(0−h)2+(0−k)2=r2

⇒h2+k2=r2

∴ equation of the circle is

(x−h)2+(y−k)2=h2+k2

It is given that the circle makes intercepts a and b on the coordinate axes.

That is the circle passes through the points (a, 0) and (0, b)

∴ Equation of the circle is

(a−h)2+(0−k)2=h2+k2

(i.e) a2−2ah+h2+k2=h2+k2

⇒a2−2ah=0

(i.e) a(a−2h)=0

⇒a=0 or a=2h

Similarly

Equation of the circle when it passes through (0,b) is

(0−h)2+(b−k)2=h2+k2

⇒h2+b2−2bk+k2=h2+k2

(i.e) b2−2bk=0

⇒b(b−2k)=0

⇒b=0 or b=2k

But a≠0 and b≠0

∴a=2h and b=2k

or h=a2 and k=b2

∴ Equation of the circle is

(x−a2)2+(y−b2)2=(a2)2+(b2)2

On expanding we get,

x2−ax+a24+y2−by+b24=a24+b24

⇒x2+y2−ax−by=0

is the equation of the required circle.