`A (4,0), B(-4,0)` are two points. The locus of `P` which moves such that `|PA-PB| = 6` is (i)`9x^2-7y^2+63=0` (ii)`9x^2+7y^2-63=0` (iii)`9x^2+7y^2+63=0` (iv)`9x^2-7y^2-63=0`

(i) x^(2) - 4x + 3 = 0 (ii) y^(2) - 7y+10 = 0

(i) x^(2)+9x+20=0 (ii) y^(2)+7y+12=0

The locus of the point P which is equidistant from 3x+4y+5=0 and 9x + 12y +7=0

The equation of the hyperbola with centre at the origin the length of transverse axis 6 and one focus is (0,4) is (i) 9x^(2) - 7y^(2) = 63 (ii) 7y^(2) - 9x^(2) = 63 (iii) 9y^(2) - 7x ^(2) = 63 (iv) 7x^(2)-9y^(2)=63

The locus of the point p which is equidstant from 3x+4y+5=0 and 9x+12y+7=0, is

The equation of the line through (-1,5) making an intercept of -2 on y-axis is (i) x+7y+2=0 (ii) 7x+y+2=0 (iii) x-7y+2=0 (iv) 7x-y+2=0

If A=(0,4),B=(0,-4) and |PA-PB|=6 and the locus of P is 7y^(2)-9x^(2)=9k ,Then k is

(6,0),(0,6),a n d(7,7) are the vertices of a A B C . The incircle of the triangle has equation. (a) x^2+y^2-9x-9y+36=0 (b) x^2+y^2+9x-9y+36=0 (c) x^2+y^2+9x+9y-36=0 (d) x^2+y^2+18 x-18 y+36=0

Calculate the value of x in the following equation :- (i) x^(2)-5x+6=0 (ii) x^(2)-9x+20=0 (iii) x^(2)-15x+56=0 (iv) x^(2)-16x+63=0

Calculate the value of x in the following equation :- (i) x^(2)-5x+6=0 (ii) x^(2)-9x+20=0 (iii) x^(2)-15x+56=0 (iv) x^(2)-16x+63=0