If a circle C_1: x^2+y^2=16 intersects anothe | vedclass

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(A) Given circle $C_1: x^2+y^2=16$ has radius $r_1=4$ and centre $O_1(0,0)$.The maximum length of a common chord is the diameter of the smaller circle

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given circle $C_1: x^2+y^2=16$ has radius $r_1=4$ and centre $O_1(0,0)$.The maximum length of a common chord is the diameter of the smaller circle,which is $2r_1 = 8$ units. This chord must pass through the centre $O_1(0,0)$ of $C_1$.The equation of the chord passing through $(0,0)$ with slope $m = \frac{3}{4}$ is $y = \frac{3}{4}x$,or $3x - 4y = 0$.Let the centre of circle $C_2$ be $O_2(h, k)$. Since the common chord is a diameter of $C_1$,the line joining the centres $O_1O_2$ is perpendicular to the common chord.The slope of the common chord is $\frac{3}{4}$,so the slope of the line $O_1O_2$ is $-\frac{4}{3}$.Thus,the coordinates of $O_2$ can be represented as $(3a, -4a)$ for some constant $a$.The distance from $O_1(0,0)$ to the chord is $0$. The distance from $O_2(3a, -4a)$ to the chord $3x - 4y = 0$ is $d = \frac{|3(3a) - 4(-4a)|}{\sqrt{3^2 + (-4)^2}} = \frac{|9a + 16a|}{5} = \frac{|25a|}{5} = 5|a|$.In the right-angled triangle formed by the radius of $C_2$ $(R_2=5)$,the distance $d$,and half the chord length $(4)$,we have $R_2^2 = d^2 + 4^2$,so $25 = d^2 + 16$,which gives $d^2 = 9$,so $d = 3$.Therefore,$5|a| = 3 \Rightarrow |a| = \frac{3}{5}$.If $a = \frac{3}{5}$,$O_2 = \left(3 \cdot \frac{3}{5}, -4 \cdot \frac{3}{5}\right) = \left(\frac{9}{5}, -\frac{12}{5}\right)$.If $a = -\frac{3}{5}$,$O_2 = \left(3 \cdot -\frac{3}{5}, -4 \cdot -\frac{3}{5}\right) = \left(-\frac{9}{5}, \frac{12}{5}\right)$.Comparing with the options,$\left(-\frac{9}{5}, \frac{12}{5}\right)$ is option $A$.

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

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