Question

es both the
The equation of the circle which touches x-axis at (0,0)
and touches the line 3x + 4y - 5 = 0 is
makes one​

Answer 1

EXPLANATION.

$\sf : \implies \:equation \: of \: the \: circle \: which \: touches \: x - axis \: (0,0) \\ \\ \sf : \implies \: touches \: the \: line \: 3x + 4y - 5 = 0 \\ \\ \sf : \implies \: centre \: of \: the \: circle \: lies \: on \: y - \: axis = (0,k)$

$\sf : \implies \: equation \: of \: circle \: = {x}^{2} + {y}^{2} = {a}^{2} \\ \\ \: \sf : \implies \: (x - 0) {}^{2} + (y - k) {}^{2} = {k}^{2} \\ \\ \sf : \implies \: \: {x}^{2} + {y}^{2} + {k}^{2} - 2ky = {k}^{2} \\ \\ \sf : \implies \: {x}^{2} + {y}^{2} - 2ky \: = 0 \: \: .....(1)$

$\sf : \implies \: circle \: touch \: the \: line \: 3x + 4y - 5 = 0 \\ \\ \sf : \implies \: \: k \: = | \frac{3(0) + 4(k) - 5}{ \sqrt{ {3}^{2} + {4}^{2} } } | \\ \\ \sf : \implies \: k \: = | \frac{4k - 5}{5} | \\ \\ \sf : \implies \: 5k \: = 4k - 5 \\ \\ \sf : \implies \: k = - 5$

$\sf : \implies \: equation \: of \: circle \\ \\ \sf : \implies \: {x}^{2} + {y}^{2} - 2ky \: = 0 \\ \\ \sf : \implies \: {x}^{2} + {y}^{2} - 2( - 5)y = 0 \\ \\ \sf : \implies \: {x}^{2} + {y}^{2} + 10y = 0$

Answer 2

$\bf{\gray{\underbrace{\blue{GIVEN:-}}}}$

• The Equation of a circle touches x-axis at (0 , 0) .

• And touches the line "3x + 4y - 5 = 0" .

$\bf{\gray{\underbrace{\blue{TO\:FIND:-}}}}$

• The Equation of circle .

$\bf{\gray{\underbrace{\blue{SOLUTION:-}}}}$

Let,

• The Equation of the circle touches y-axis at (0 , k)

✨ So, Radius of the circle is "k" .

• $\bf\red{Center}$ = (h , k) = (0 , k)

☯︎ Firstly, we can find the perpendicular distance of the straight line 3x + 4y - 5 = 0 from center (h , k) = (0 , k) by using the equation;

$\orange\bigstar\:\bf{\gray{\overbrace{\underbrace{\purple{Distance\:=\:\dfrac{\left|\:ah\:+\:bk\:+\:c\:\right|}{\sqrt{a^2\:+\:b^2}}\:}}}}}$

✨ The value of a, b and c can be obtained from 3x + 4y - 5 = 0;

• $\bf\red{a}$ = 3

• $\bf\red{b}$ = 4

• $\bf\red{c}$ = -5

$\rm{:\implies\:Distance\:=\:\dfrac{\left|\:3\times{0}\:+\:4\times{k}\:-\:5\:\right|}{\sqrt{3^2\:+\:4^2}}\:}$

• $\bf\red{Distance}$ = Radius = k

$\rm{:\implies\:k\:=\:\dfrac{\left|\:0\:+\:4k\:-\:5\:\right|}{\sqrt{9\:+\:16}}\:}$

$\rm{:\implies\:k\:=\:\dfrac{\left|\:4k\:-\:5\:\right|}{\sqrt{25}}\:}$

$\rm{:\implies\:k\:=\:\dfrac{\left|\:4k\:-\:5\:\right|}{5}\:}$

$\rm{:\implies\:5k\:=\:4k\:-\:5\:}$

$\rm{:\implies\:5k\:-\:4k\:=\:-5\:}$

$\bf\green{:\implies\:k\:=\:-5\:}$

• $\bf\red{(h\:,\:k)}$ = (0 , -5)

☯︎ Now for equation of the circle, we can use the following equation :-

$\green\bigstar\:\bf{\gray{\overbrace{\underbrace{\purple{(x\:-\:h)^2\:+\:(y\:-\:k)^2\:=\:(Radius)^2\:}}}}}$

• $\bf\red{Radius}$ = -5

$\rm{:\implies\:(x\:-\:0)^2\:+\:\Big\{y\:-\:(-5)\Big\}^2\:=\:(-5)^2\:}$

$\rm{:\implies\:x^2\:+\:(y\:+\:5)^2\:=\:25\:}$

$\rm{:\implies\:x^2\:+\:y^2\:+\:\cancel{25}\:+\:10y\:=\:\cancel{25}\:}$

$\bf\blue{:\implies\:x^2\:+\:y^2\:+\:10y\:=\:0\:}$

$\red\therefore$ The Equation of circle is " x² + y² + 10y = 0 " .