Equation of Circle Whose Extremities of a Diameter Given

Q. If one end of a diameter of the circle $x^2+y^2-4x-6y+11=0$ be (3, 4), then the other end is

1. (0, 0)
2. (1, 1)
3. (1, 2)
4. (2, 1)

Q.

A line cuts the x-axis at A(4, 0) and the y-axis at B(0, 8). A variable line PQ is drawn perpendicular to AB cutting the x-axis in P and the y-axis in Q. If AQ and BP intersect at R, find the locus of R.

1. $x^2+y^2-2x-4y=0$

2. $x^2+y^2+2x+4y=0$

3. $x^2+y^2-2x+4y=0$

4. $x^2+y^2-4x-8y=0$

Q. The equation of circle whose diameter is the line joining the points (­–4, 3) and (12, –1) is

1. $x_2+(y_2+8x+2y+51=0$
2. $x_2+(y_2+8x-2y-51=0$
3. $x_2+(y_2+8x+2y-51=0$
4. $x_2+(y_2-8x-2y-51=0.$

Q. If the line $y=4x-2$ cuts the curve $y^2=8x$ at points $A$ and $B,$ then the equation of circle having $AB$ as a diameter, is

1. ${(x+\frac{1}{2})}^2+y^2-\frac{x}{2}-2y-4=0$
2. ${(x-\frac{1}{2})}^2+y^2-\frac{x}{2}-2y-4=0$
3. ${(x-\frac{1}{2})}^2+y^2+\frac{x}{2}-2y-4=0$
4. ${(x-\frac{1}{2})}^2+y^2-\frac{x}{2}+2y-4=0$