Coordinate Geometry Set-1

$(4,1)$

$(5,-1)$

$(5,1)$

$(6,1)$

$y=x-3$

$y=-x-3$

$y=x+3$

$y=-x+3$

$10$

$\tfrac{11}{2}$

$21$

$\tfrac{21}{2}$

Both A and R are true but R is not a correct explanation of A.

Both A and R are true and R is a correct explanation of A.

A is true and R is false.

A is false and R is true.

No such circle exists

$(x-4)^2+\Bigl(y-\dfrac{13}{4}\Bigr)^2=\dfrac{425}{16}$

$(x-4)^2+\Bigl(y-\dfrac{17}{4}\Bigr)^2=\dfrac{289}{16}$

$x^2+y^2-4x-\dfrac{19}{2}y=0$

$5\sqrt{5}$

$5\sqrt{2}$

$4\sqrt{2}$

$5\sqrt{3}$

$y-5=-\dfrac{2}{3}(x-5)$

$2x+3y-25=0$

$y=\dfrac{2}{3}x+\dfrac{1}{3}$

$3x+2y-25=0$

$(2,\,3)$

$\left(\dfrac{3}{2},\dfrac{5}{2}\right)$

$(4,\,5)$

$\left(\dfrac{7}{4},\dfrac{9}{4}\right)$

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

$(x-1)^2+(y-1)^2=\dfrac{1}{2}$

$\left(x-\dfrac{3}{2}\right)^2+\left(y-\dfrac{3}{2}\right)^2=\dfrac{5}{2}$

$x^2+y^2-4x-4y+5=0$

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

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

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

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

$y=-x+4$

$y=x-2$

$y=x+2$

$y=-x+2$

$y=\left(-3+\dfrac{3\sqrt{3}}{2}\right)x+9+3\sqrt{3}\quad\text{and}\quad y=\left(-3-\dfrac{3\sqrt{3}}{2}\right)x+9-3\sqrt{3}$

$y=\left(3+\dfrac{3\sqrt{3}}{2}\right)x-9+3\sqrt{3}\quad\text{and}\quad y=\left(3-\dfrac{3\sqrt{3}}{2}\right)x-9-3\sqrt{3}$

$y=\left(-\dfrac{3}{2}+\dfrac{\sqrt{3}}{2}\right)x+3-\sqrt{3}\quad\text{and}\quad y=\left(-\dfrac{3}{2}-\dfrac{\sqrt{3}}{2}\right)x+3+\sqrt{3}$

$y=\left(-3+\dfrac{3\sqrt{3}}{2}\right)x+9-3\sqrt{3}\quad\text{and}\quad y=\left(-3-\dfrac{3\sqrt{3}}{2}\right)x+9+3\sqrt{3}$

$(0,3)$

$(2,1)$

$(4,1)$

$(1,2)$

$(x+ \tfrac{13}{4})^2+y^2=\left(\tfrac{13}{4}\right)^2$

$(x-5)^2+y^2=25$

$\left(x-\tfrac{13}{4}\right)^2+y^2=\left(\tfrac{13}{4}\right)^2$

$(x-2)^2+(y-3)^2= \left(\tfrac{13}{4}\right)^2$

$\left(\dfrac{7}{3}, -\dfrac{2}{3}\right)$

No point exists on $x+2y-3=0$ that is equidistant from $A$ and $B$

$\left(2, \dfrac{1}{2}\right)$

$\left(0, \dfrac{3}{2}\right)$