Equation of the locus of all points such that the difference of its distances from `(-3,-7), (-3,3)` is `8` is (A) `(x+3)^2/16-(y+2)^2/9=1` (B) `(x+3)^2/9-(y+2)^2/16=-1` (C) `(x+3)^2/9-(y+2)^2/19=1` (D) `(x+3)^2/7-(y+2)^2/19=-1`

Equation of the locus of all points such that the difference of its distances from (-3,-7),(-3,3) is 8 is (A) ((x+3)^(2))/(16)-((y+2)^(2))/(9)=1(B)((x+3)^(2))/(9)-((y+2)^(2))/(16)=1(C)((x+3)^(2))/(7)-((y+2)^(2))/(19)=1(D)((x+3)^(2))/(7)-((y+2)^(2))/(19)=-1

The equation of the locus of points which are equidistant from the points (2,-3) and (3,-2) is (A) x+y=0 (B) x+y=7 (C) 4x+4y=38 (D) x+y=1

The equation of the ellipse whose vertices are (9,2),(-1,2) and the distance between the foci is 8 unit is (A)9(x-3)^(2)+25(y-4)^(2)=225(B)9(x+3)^(2)+25(y+4)^(2)=225(C)9(x-4)^(2)+25(y-3)^(2)=225(D)9(x+4)^(2)+25(y+3)^(2)=225

Find the centre and radius of each of the following circles : (i) (x - 3)^(2) + (y- 1) ^(2) = 9 (ii) (x - (1)/(2) ) ^(2) + ( y + (1)/(3) ) ^(2) = (1)/(16) (iii) (x + 5) ^(2) + ( y- 3 ) ^(2) = 20 (iv) x ^(2) + (y- 1 ) ^(2) = 2

Let vertices of the triangle ABC is A(0,0),B(0,1) and C(x,y) and perimeter is 4 then the locus of C is : (A)9x^(2)+8y^(2)+8y=16(B)8x^(2)+9y^(2)+9y=16(C)9x^(2)+9y^(2)+9y=16(D)8x^(2)+9y^(2)-9x=16

The number of real points are which the line 3y-2x=14 cuts the hyperbola ((x-1)^(2))/(9)+((y-2)^(2))/(4)=5 is (A) 1(B)2(C)3 (D) 4

Locus of feet of perpendicular from (5,0) to the tangents of (x^(2))/(16)-(y^(2))/(9)=1 is (A)x^(2)+y^(2)=4(B)x^(2)+y^(2)=16(C)x^(2)+y^(2)=9(D)x^(2)+y^(2)=25