The locus of centre of circles which cuts orthogonally the circle x2 + y2 - 4x + 8 = 0 and touches x + 1 = 0, is
y2 + 6x + 7 = 0
x2 + y2 + 2x + 3 = 0
x2 + 3y + 4 = 0
None of the above
A.
y2 + 6x + 7 = 0
x2 + y2 + 2gx + 2fy + c = 0
Given equation of circle is
x2 + y2 - 4x + 8 = 0
The centres of above circles are (- g, - f) and (2, 0)
Condition of orthogonality is
2(g1g2 + f1f2) = c1 + c2
Also, the assume circle touch the lme x + 1 = 0.
The perpendicular drawn from centre to the line is equal to radius
On squaring both sides, we get
Putting the value of c in Eq (i), we get
4g = f2 + 2 - 1 + 8
Thus, locus of circle is y2 + 6x + 7 = 0.
The least value of a, for which the function a has at east one solution in the interval , is
9
4
5
1
If one line ofregression coefficient is less than unity, then the other will be
less than unity
equal to unity
greater than unity
All of these