Question

$y=x^2-2x-15 \\ y=2x+6$

Which of the following could be the y-coordinate of a point of intersection of the graphs of the two equations above in the xy-plane?

A) -3
B) 5
C) 7
D) 20​

Answer 1

$\large\underline{\sf{Solution-}}$

The given equations are

$\: \: \: \: \: \: \: \: \: \: \: \: \bf \: y \: = {x}^{2} - 2x - 15 - - - - (1)$

and

$\: \: \: \: \: \: \: \: \: \: \: \: \bf \: y \: = 2x + 6 - - - (2)$

For the point of intersection of equation (1) and (2),

we have to equate the equations.

So,

$\rm :\longmapsto\: {x}^{2} - 2x - 15 = 2x + 6$

$\rm :\longmapsto\: {x}^{2} - 4x - 21 = 0$

$\rm :\longmapsto\: {x}^{2} - 7x + 3x - 21 = 0$

$\rm :\longmapsto\:x(x - 7) + 3(x - 7) = 0$

$\rm :\longmapsto\:(x - 7)(x + 3) = 0$

$\bf\implies \:x = 7 \: \: \: \: or \: \: \: \: x = - 3$

So, corresponding values of y is

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y = 2x + 6 \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 7 & \sf 20 \\ \\ \sf - 3 & \sf 0 \end{array}} \\ \end{gathered}$

$\: \: \: \: \: \boxed{ \bf \:Hence, \: option \: (D) \: is \: correct.}$

Answer 2

Answer:

Equation of the circle with center (0,1) & radius 3 is given as

(x−0)2+(y−1)2=32⟹x2+y2−2y−8=0