If the circle {C_1}: {x^2} + {y^2} = 16 inter | vedclass

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

Question

(A) Let $(h, k)$ be the coordinates of the centre of circle ${C_2}$.Its equation is $(x - h)^2 + (y - k)^2 = 5^2$.The equation of ${C_1}$ is $x^2 + y^

Vedclass · Accepted Answer

$\left( -\frac{9}{5}, \frac{12}{5} \right), \left( \frac{9}{5}, -\frac{12}{5} \right)$

Vedclass · Answer

(A) Let $(h, k)$ be the coordinates of the centre of circle ${C_2}$.Its equation is $(x - h)^2 + (y - k)^2 = 5^2$.The equation of ${C_1}$ is $x^2 + y^2 = 4^2$. The equation of the common chord of ${C_1}$ and ${C_2}$ is $2hx + 2ky = h^2 + k^2 - 9$ $(i)$.Let $p$ be the length of the perpendicular from the centre $(0, 0)$ of ${C_1}$ to $(i)$.Then $p = \frac{|h^2 + k^2 - 9|}{\sqrt{4h^2 + 4k^2}}$.The length of the common chord is $2\sqrt{4^2 - p^2}$,which is of maximum length if $p = 0$.This implies $h^2 + k^2 - 9 = 0$ $(ii)$.The slope of $(i)$ is given as $\frac{3}{4}$.Thus,$-\frac{h}{k} = \frac{3}{4} \Rightarrow k = -\frac{4h}{3}$ $(iii)$.Substituting $(iii)$ into $(ii)$,we get $h^2 + (-\frac{4h}{3})^2 = 9$ $\Rightarrow h^2 + \frac{16h^2}{9} = 9$ $\Rightarrow \frac{25h^2}{9} = 9$ $\Rightarrow h^2 = \frac{81}{25}$ $\Rightarrow h = \pm \frac{9}{5}$.If $h = \frac{9}{5}$,then $k = -\frac{4}{3} 	imes \frac{9}{5} = -\frac{12}{5}$.If $h = -\frac{9}{5}$,then $k = -\frac{4}{3} 	imes (-\frac{9}{5}) = \frac{12}{5}$.Thus,the centres are $\left( -\frac{9}{5}, \frac{12}{5} ight)$ and $\left( \frac{9}{5}, -\frac{12}{5} ight)$.

If the circle ${C_1}: {x^2} + {y^2} = 16$ intersects another circle ${C_2}$ of radius $5$ in such a manner that the common chord is of maximum length and has a slope equal to $\frac{3}{4}$,the coordinates of the centre of ${C_2}$ are

Similar Questions

The length of the chord joining the points in which the straight line $\frac{x}{3} + \frac{y}{4} = 1$ cuts the circle ${x^2} + {y^2} = \frac{169}{25}$ is

The equation of the common tangent to the circles $x^2+y^2-4x+10y+20=0$ and $x^2+y^2+8x-6y-24=0$ is

If the equation of the circle having the common chord of the circles $x^2+y^2+x-3y-10=0$ and $x^2+y^2+2x-y-20=0$ as its diameter is $x^2+y^2+\alpha x+\beta y+\gamma=0$,then $\alpha+2\beta+\gamma=$

If the circles $x^2+y^2-2x-2y-7=0$ and $x^2+y^2+4x+2y+k=0$ cut orthogonally,then the length of their common chord is units.

The length of the common chord of the circles $x^2 + y^2 - 6x + 5 = 0$ and $x^2 + y^2 - 2y - 3 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module