Question

Find the equation of a circle which pass through origin and cut off intercept equal to 3&4

Answer 1

Answer:

Let the equation of a circle be

x^2 +y^2 +2gx+2fy+c=0

The circle is passing through origin then c=0

X-intercept=3

2v(g^2 - c) =3

+-(2g)=3

g=+-3/2

Similarly f =+-2

Therefore the equation of a circle be x^2 +y^2 +-3x+-4y=0

Answer 2

Corrected question :

• Find the equation of a circle which pass through origin and cut off intercepts equal to 3 and 4 from the positive parts or the axes respectively.

Required Solution :

Here we know that intercept of length 4 is being intercepted from y-axis and intercept of length of 3 is intercepted from x-axis and the circle is passing from origin.

So we would be giving notations now !

• Assuming the second point from which circle has intercepted in y-axis as "B" (First is origin)
• Assuming the second point from which circle has intercepted in x-axis as "A" (First is origin)
• Point in x-axis till centre as "L"
• Centre of circle as "C"

~ Distance between OB is of 4 units

~ Distance between OL is of 3/2 units

~ Distance between OA is of 3 units

So we need to calculate now the distance between OC (That is the radius of circle). So we would be using Pythagoras theorem inorder to do so (As OCL is a ∆).

→ (Hypotenuse)² = (Base)² + (Perpendicular)²

→ (OC)² = (OL)² + (LC)²

→ OC² = (3/2)² + (2)²

→ OC² = (9/4) + (4)

→ OC² = 25/4

→ OC = √(25 / 4)

→ OC = 5/2

Therefore,

• Radius is of 5/2 units.

As we know equation of circle :

• (x - h)² + (y - k)² = r²

Here,

• h and k are coordinates of centre.

→ (x - 3/2)² + (y - 2)² = (5/2)²