Question

If a circle passes through the point `(a, b)` and cuts the circle `x^2 + y^2 = 4` orthogonally, then the locus of its centre is
A. `2ax+2by-(a^(2)+b^(2)+k^(2))=0`
B. `2ax+2by-(a^(2)-b^(2)+k^(2))=0`
C. `x^(2)+y^(2)-3ax-4by+(a^(2)+b^(2)-k^(2))=0`
D. `x^(2)+y^(2)-2ax-3by+(a^(2)-b^(2)-k^(2))=0`

Answer 1

Correct Answer - 1
Let the equation of the circle through (a,b) be
`x^(2)+y^(2)+2gx+2fy+c=0` (1)
SO, `a^(2)+b^(2)+2ag+2bf+c=0` (2)
Since circle (1) cuts `x^(2)+y^(2)=k^(2)` orthogonally, we have
`2g(0)+2f(0)=c-k^(2)` or `c=k^(2)`
Putting `c=k^(2)` in (2) ,we get
`2ag+2bf+(a^(2)+b^(2)+k^(2))=0`
So, the locus of the center `(-g,-f)` is
`-2ax-2by+(a^(2)+b^(2)+k^(2))=0` ltbrlt or `2ax+2by- (a^(2)+b^(2)+k^(2))=0`
Alternate Methode `:`

As shown in the figure, circle `C_(20` intersect the circle `C_(1) : x^(2)+y^(2)= k^(2)` orthogonally at point Q.
Circle `C_(2)` passes through the point P(a,b) . We have to fnd the locus of center `R ( alpha, beta ) ` of circle `C_(2)`
Now from the figure, `RQ=P=` radiua of the circle `C_(2)`
`:. RP^(2) = RQ^(2)`
`:. RP^(2)= or^(2) - OQ^(2)`
`:. (alpha -a)^(2) + ( beta -b)^(2) = (alpha^(2)+beta^(2)) - k^(2)`
`:. ` locus is `2ax +2by -(a^(2)+b^(2)+k^(2)) =0`