If the coordinates of the point of contact of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The circle $x^2+y^2-4x+8y+4=0$ has center $P(2, -4)$ and radius $r_1 = \sqrt{2^2+(-4)^2-4} = \sqrt{4+16-4} = 4$. The circle $x^2+y^2+2x=0$ has cen

Vedclass · Accepted Answer

-$2$

Vedclass · Answer

(B) The circle $x^2+y^2-4x+8y+4=0$ has center $P(2, -4)$ and radius $r_1 = \sqrt{2^2+(-4)^2-4} = \sqrt{4+16-4} = 4$. The circle $x^2+y^2+2x=0$ has center $Q(-1, 0)$ and radius $r_2 = \sqrt{(-1)^2+0^2-0} = 1$. Since the circles touch externally,the point of contact $O(a, b)$ divides the line segment $PQ$ internally in the ratio $r_1 : r_2 = 4 : 1$. Using the section formula,$O(a, b) = \left( \frac{4(-1)+1(2)}{4+1}, \frac{4(0)+1(-4)}{4+1} \right)$. $O(a, b) = \left( \frac{-4+2}{5}, \frac{0-4}{5} \right) = \left( -\frac{2}{5}, -\frac{4}{5} \right)$. Thus,$a = -\frac{2}{5}$ and $b = -\frac{4}{5}$. Therefore,$a+2b = -\frac{2}{5} + 2\left( -\frac{4}{5} \right) = -\frac{2}{5} - \frac{8}{5} = -\frac{10}{5} = -2$.

If the coordinates of the point of contact of the circles $x^2+y^2-4x+8y+4=0$ and $x^2+y^2+2x=0$ are $(a, b)$,then $a+2b=$

Similar Questions

The equation of the circle having the chord $x \cos \alpha + y \sin \alpha = p$ of the circle $x^2 + y^2 = a^2$ as its diameter is:

The condition for the circles $x^2+y^2+ax+4=0$ and $x^2+y^2+by+4=0$ to touch each other is

If $x-y+1=0$ meets the circle $x^2+y^2+y-1=0$ at $A$ and $B$,then the equation of the circle with $AB$ as diameter is

The centre of the smallest circle which cuts the circles $x^2+y^2-2x-4y-4=0$ and $x^2+y^2-10x+12y+52=0$ orthogonally is

The limiting points of the co-axial system containing the two circles $x^2+y^2+2x-2y+2=0$ and $25(x^2+y^2)-10x-80y+65=0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module