`(ax)/b-(by)/a=a+b`, `ax-by=2ab`

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)+4)=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

A, B C and D are the points of intersection with the coordinate axes of the lines ax+by=ab and bx+ay=ab, then

Solve : (a)/(ax-1)+(b)/(bx-1)=a+b , where a+b ne 0, ab ne 0 .

{:(ax + by = 1),(bx + ay = ((a + b)^(2))/(a^(2) + b^(2))-1):}

If a-b,ax-by,ax^2-by^2a,b!=0) are in G.P., then x,y (ax-by)/(a-b) are in (A) A.P. only (B) G.P.only (C) A.P., G.P. (D) A.P., and G.P and H.P

If (ax)/(costheta)+(by)/(sintheta)=a^2-b^2 , and (axsintheta)/(cos^2theta)-(bycostheta)/(sin^2theta)=0 , prove that (ax)^(2/3)+(by)^(2/3)=(a^2-b^2)^(2/3)

The two lines ax+by+c=0 and a'x+b'y+c'=0 are perpendicular if (i) ab'=a'b (ii) ab+a'b'=0 (iii) ab'+a'b=0 (iv) a a'+b b'=0

The value of |{:(x^(2)+y^(2),ax+by,x+y),(ax+by,a^(2)+b^(2),a+b),(x+y,a+b,2):}| depends on

if one of the lines given by the equation ax^2+2hxy+by^2=0 coincides with one of the lines given by a'x^2+2h'xy+b'y^2=0 and the other lines representted by them be perpendicular , then . (ha'b')/(b'-a')=(h'ab)/(b-a)=1/2sqrt((-aa'bb') .

If x^2+px+1 is a factor of ax^3+bx+c then a) a^2+c^2=-ab b) a^2+c^2ab c) a^2-c^2=ab d) a^2-c^2=-ab