[" 41."(x)/(a)+(y)/(b)=2],[qquad ax-by=a^(2)-b^(2)]

Solve: (x)/(a)+(y)/(b)=2,quad ax-by=a^(2)-b^(2)

Solve for xandy: (x)/(a)+(y)/(b)=2,ax-by=a^(2)-b^(2)

x+y=a+b,ax-by=a^(2)-b^(2)

(x)/(a)-(y)/(b)=0,ax+by=a^(2)+b^(2)

The value of y which satisfies (x)/(a)=(y)/(b) and ax+by=a^(2)+b^(2) is

{:(x + y = a + b),(ax - by = a^(2) - b^(2)):}

The locus of the midpoints of the chords of the circle x^(2)+y^(2)-ax-by=0 which subtend a right angle at ((a)/(2),(b)/(2)) is ax+by=0ax+by=a^(2)=b^(2)x^(2)+y^(2)-ax-by+(a^(2)+b^(2))/(8)=0x^(2)+y^(2)-ax-by-(a^(2)+b^(2))/(8)=0

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

If (x+y)/(3a-b) = (y+z)/(3b-c) = (z+x)/(3c-a) , then show that (x+y+z)/(a+b+c)= (ax + by+cz)/(a^(2)+b^(2)+c^(2))