Question

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

• A. $\frac{-5}{2}$
• B. $\frac{-7}{2}$
• C. $\frac{-3}{2}$
• D. $\frac{5}{2}$
• E. $\frac{7}{2}$

Answer 1

The correct option is B $\frac{-7}{2}$

Equation of the circle is $x^2+y^2-8x+2y=0$ or $(x-4{)}^2+(y+1{)}^2=17$.

Hence, coordinates of the centre of the circle are $(4,-1)$

Equation of the parabola is $y=x^2-4x+10$ or $(y-6)=(x-2{)}^2$.

Hence, coordinates of the vertex of the parabola are $(2,6)$

Therefore, slope of the line joining centre of the circle to the vertex of the parabola $=\frac{6-(-1)}{2-4}=\frac{-7}{2}$.