The slope of the straight line joining the centre of the circle x2 + y2-8x + 2y = 0 and the vertex of the parabola y = x2 - 4x + 10 is

Question

The slope of the straight line joining the centre of the circle x2 + y2-8x + 2y = 0 and the vertex of the parabola y = x2 - 4x + 10 is (A) (5/2) (B) (7/2) (C) (3/2) (D)  (9/2)

Tardigrade Admin · Accepted Answer

Centre of the circle x2+y2-8x+2y =0 is (4, -1) Given parabola is y= (x-2)2 +6 ⇒ (x-2)2=4⋅(1/4)(y-6) Vertex is (2, 6) Required slope = ((6/2) (1/4)) =- (7/2)

Q. The slope of the straight line joining the centre of the circle $x^{2} + y^{2} - 8 x + 2 y = 0$ and the vertex of the parabola $y = x^{2} - 4 x + 10$ is

$\frac{5}{2}$

$\frac{7}{2}$

$\frac{3}{2}$

$\frac{9}{2}$

Centre of the circle $x^{2} + y^{2} - 8 x + 2 y = 0$ is $(4, - 1)$
Given parabola is $y = (x - 2)^{2} + 6$
$\Rightarrow (x - 2)^{2} = 4 \cdot \frac{1}{4} (y - 6)$
Vertex is $(2, 6)$
Required slope $= (\frac{6}{2} \frac{1}{4})$
$= - \frac{7}{2}$

Q. The slope of the straight line joining the centre of the circle x2+y2−8x+2y=0 and the vertex of the parabola y=x2−4x+10 is

25​

27​

23​

29​