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# A circle passing through (0,0), (2,6), (6,2) cuts the x axis a the point P ? (0,0). Then the length of OP, where O is origin,is

Arun
25763 Points
11 months ago
Let x2+y2+2gx+2fy=0 be the equation of circle passing through origin
Given that it also passes through (2,6) and (6,2)
4+36+4g+12f=0
g+3f+10=0–eq.1
And
36+4+4f+12fg=0
f+3g+10=0–eq.2
Solving eq.1 and eq.2
f–g=2−5​
Equation is 2x2+2y2–10x−10y=0
Substituting y=0
2x2–10x=0⟹x=5,0
P=(5,0),O=(0,0)
OP=5
Vikas TU
14149 Points
11 months ago
Dear student
The above equation is wrongly typed.
Let x^2+y^2+2gx+2fy=0 be the equation of circle passing through origin
Given that it also passes through (2,6) and (6,2)
4+36+4g+12f=0
g+3f+10=0–eq.1
And
36+4+4f+12fg=0
f+3g+10=0–eq.2
Solving eq.1 and eq.2
f–g=2−5​
Equation is 2x^2+2y^2–10x−10y=0
Substituting y=0
2x^2–10x=0
⟹x=5,0
P=(5,0),O=(0,0)
OP=5