The circle x2+y2−6x−6y+9=0 is inscribed in a triangle which has two of its sides along the coordinate axes. The locus of the circumcentre of the triangle is x+y−ba+bxy+a(x2+y2)1b=0. Find a+b
A
3.0
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B
3.00
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C
3
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Solution
Equation of the given circle (x−3)2+(y−3)2=9
Let the equation of the third side AB is xm+yn=1
Length of the perpendicular from C=3 ∴∣∣3m+3n−1∣∣√1m2+1n2=3 ∵ origin and C lie on the same side of AB ⇒3m+3n−1√1m2+1n2=−33m+3n−1=−3√1m2+1n2...(1) ∵∠AOB=π2 ∴AB will be the diameter of the circle which touches AB at K ∴ centre of the circle is (m2,n2)
Let the locus of circumcenter be (h,k)≡(m2,n2)
Now from (1), 32h+32k−1=−3√1(2h)2+1(2k)2⇒3h+3k−2hk=−3√h2+k2 ∴ Locus of the centre (h,k) is x+y−2xy3+√x2+y2=0