Question

If the line x + 2y = 3 cuts a chord of length r unit with the arcle x2 + y2 =r2 then find r2

Answer 1

ANSWER.

Option [ A ] is correct.

EXPLANATION.

$\sf \to \: if \: the \: line \: = x + 2y = 3 \: cuts \: the \: chord \: r \: unit \: \\ \\ \sf \to \: circle \: = {x}^{2} + {y}^{2} = {r}^{2}$

To find the value of r².

$\sf \to \: lines \: = x + 2y - 3 = 0 \\ \\ \sf \to \:it \: cuts \: the \: chord \: = r \: units \\ \\ \sf \to \: half \: of \: the \: chord \: = \dfrac{r}{2} = base \\ \\ \sf \to \: hypotenuse \: = r \\ \\ \sf\to to \: \: find \: perpendicular$

$\sf \to \: by \: using \: perpendicular \: distance \: formula \\ \\ \sf \to \: p = | \frac{(0) + 2(0) - 3}{ \sqrt{ {1}^{2} + {2}^{2} } } | \\ \\ \sf \to \: p \: = \dfrac{3}{ \sqrt{5} } \\ \\ \sf \to \: by \: using \: pythagorus \: theorm \\ \\ \sf \to \: {h}^{2} = {p}^{2} + {b}^{2} \\ \\ \sf \to \: {r}^{2} = \frac{(3) {}^{2} }{ (\sqrt{5} \: ) {}^{2} } + \dfrac{ {r}^{2} }{4} \\ \\ \sf \to {r}^{2} = \frac{9}{5} + \frac{ {r}^{2} }{4} \\ \\ \sf \to \: 20 {r}^{2} = 36 + 5 {r}^{2} \\ \\ \sf \to \: 15 {r}^{2} = 36 \\ \\ \sf \to \: {r}^{2} = \frac{12}{5}$