# Q. The line `y=ax+b` intersects the curve C: `x^2+y^2+6x-10y+1=0` at the points A and B.If the line segment AB subtends a right angle at origin then the locus of the point `(a,b)` is the curve...

Q. The line `y=ax+b` intersects the curve C: `x^2+y^2+6x-10y+1=0` at the points A and B.If the line segment AB subtends a right angle at origin then the locus of the point `(a,b)` is the curve `g(x,y)=0.`

The equation of the curve `g(x,y)=0` is:-

a) `x^2+2y^2-6xy+10y+1=0`

b) `x^2+2y^2-6xy-10y+1=0`

c) `x^2-2y^2-6xy+10y+1=0`

d)`x^2-2y^2-6xy-10y+1=0`

*print*Print*list*Cite

Since the line AB subtends a right angle at the origin (0,0), then an instance of the pair (A,B) is any of the pairs of points (there are four possible pairs in fact) where the curve (a *circle*) C crosses the x-axis and y-axis respectively. These points are:

1) setting y=0 gives x= -3 +/- sqrt(8)

2) setting x=0 gives y = 5 +/- sqrt(24)

Choosing the pair A = (-3 -sqrt(8), 0) and B = (0, 5 +sqrt(24)) we now find the equation of the line y = ax + b that corresponds to the line AB.

Doing this we find that

a = (5 +sqrt(24))/(3 +sqrt(8)), b = 5 +sqrt(24)

Plug this coordinate (a,b) into the four possible equations a)-d) given.

We see that the point is consistent with equation b) only, ie subtituting x=a and y=b into the righthand side yields a value of zero, consistent with the lefthand side.

Therefore, since equation b) is consistent with a particular solution to the locus and none of the other equations are, b) must be the solution. The other three obvious possible pairs of (A,B) could be used to ascertain particular pairs (a,b) on the locus, which could then be substituted in b) to confirm this.

**Answer: equation b) describes the locus of points (a,b)**