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Circle question

2023 · 10 Apr · Shift 2 · Q31

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Mathematics
Circle
2023 · 10 Apr · Shift 2 · Q31

Circle question

2023 · 10 Apr · Shift 2 · Q31

JEE MainMathematicsCircleMCQ+4 / −1

Let A be the point

(1,2)

and B be any point on the curve

x^{2}+y^{2}=16

. If the centre of the locus of the point P, which divides the line segment

\mathrm{AB}

in the ratio

3: 2

is the point C

(\alpha, \beta)

, then the length of the line segment

\mathrm{AC}

is :

A
$\frac{3 \sqrt{5}}{5}$
B
$\frac{6 \sqrt{5}}{5}$
C
$\frac{2 \sqrt{5}}{5}$
D
$\frac{4 \sqrt{5}}{5}$

View written solutionFree

Correct answer: A

Let $A=(1,2)$ and let $B=(x,y)$ be any point on the circle $x^2+y^2=16.$
Point $P$ divides the line segment $AB$ in the ratio $3:2$ internally.

Using the section formula, if $P$ divides $AB$ in the ratio $3:2$ , then

=\left(\frac{3x+2}{5},\frac{3y+4}{5}\right).$$ Let the coordinates of $P$ be $(X,Y)$. Then $$X=\frac{3x+2}{5},\qquad Y=\frac{3y+4}{5}.$$ 3. Express $x,y$ in terms of $X,Y$: $$3x=5X-2\Rightarrow x=\frac{5X-2}{3},$$ $$3y=5Y-4\Rightarrow y=\frac{5Y-4}{3}.$$ Since $B=(x,y)$ lies on $x^2+y^2=16$, substitute: $$\left(\frac{5X-2}{3}\right)^2+\left(\frac{5Y-4}{3}\right)^2=16.$$ Multiply by $9$: $$(5X-2)^2+(5Y-4)^2=144.$$ Rewrite: $$25\left(X-\frac{2}{5}\right)^2+25\left(Y-\frac{4}{5}\right)^2=144,$$ so $$\left(X-\frac{2}{5}\right)^2+\left(Y-\frac{4}{5}\right)^2=\frac{144}{25}.$$ Thus the locus of $P$ is a circle with centre $$C=\left(\frac{2}{5},\frac{4}{5}\right).$$ Hence $$\alpha=\frac{2}{5},\qquad \beta=\frac{4}{5}.$$ 4. Now find $AC$ where $A=(1,2)$: $$AC=\sqrt{\left(1-\frac{2}{5}\right)^2+\left(2-\frac{4}{5}\right)^2}$$ $$=\sqrt{\left(\frac{3}{5}\right)^2+\left(\frac{6}{5}\right)^2}$$ $$=\sqrt{\frac{9}{25}+\frac{36}{25}}$$ $$=\sqrt{\frac{45}{25}}=\frac{\sqrt{45}}{5}=\frac{3\sqrt{5}}{5}.$$ 5. Therefore the correct option is $$\boxed{\text{A }\frac{3\sqrt{5}}{5}}.$$

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