Equation of Circle Whose Extremities of a Diameter Given

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. B & C are fixed points having co-ordinates (3, 0) and (–3, 0) respectively. If the vertical angle BAC is ${90}^{\circ }$, then the locus of the centroid of the $\Delta ABC$ has the equation.

1. 
2. 
3. 
4. 

Q.

An isosceles right angled triangle is inscribed in the circle $x^2+y^2=r^2.$ If the coordinates of an end of the hypotenuse are (a, b), the coordinates of the vertex are

1. (-a, -b)
   

2. (b, -a)
   

3. (b, a)
   

4. (-b, -a)

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. (2, 1)
4. (1, 2)

Q. Solve the linear equation:

1. $x=\frac{15}{8}$
2. $x=\frac{65}{3}$
3. $x=\frac{17}{6}$
4. $x=\frac{23}{16}$

Q. On solving $\frac{25}{x+y}-\frac{3}{x-y}=1$, $\frac{40}{x+y}+\frac{2}{x-y}=5$, we get

1. $x=8$, $y=6$
2. $x=6$, $y=4$
3. $x=4$, $y=6$
4. None of these

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

1. $x=\frac{1}{2};y=-\frac{7}{6}$
2. $x=\frac{21}{20};y=-\frac{3}{10}$
3. $x=\frac{14}{5};y=-\frac{6}{11}$
4. $x=\frac{18}{19};y=-\frac{16}{10}$

Q. The equation of a circle of radius 5 which lies within the circle $x^2+y^2+14x+10y-26=0$ and touches it at the point (-1, 3) is

1. $x^2+y^2+8x+2y-8=0$
2. $x^2+y^2+8x+2y+8=0$
3. $x^2+y^2+8x+2y-14=0$
4. None of the above

Q. The combined equation of the three sides of a triangle is $(x^2-y^2)(2x+3y-6)$. If the point $(0,a)$ lies in the interior of this triangle then

1. $0<\alpha <2$
2. $-2<\alpha <0$
3. $-2<\alpha <2$
4. $\alpha \ge 2$