Question

Q. Standard equation of the circle touching y axis at origin q

(a) x^2+y^2=r^2
(b) (x-h)^2+y^2=r^2
(c) x^2+(y-r)^2=r^2
(d) (x-r)^2+y^2=r^2​

Answer 1

SOLUTION

TO CHOOSE THE CORRECT OPTION

Standard equation of the circle touching y axis at origin

$\sf{(a) \: \: {x}^{2} + {y}^{2} = {r}^{2} }$

$\sf{(b) \: \: {(x - h)}^{2} + {y}^{2} = {r}^{2} }$

$\sf{(c) \: \: {x}^{2} + {(y - r)}^{2} = {r}^{2} }$

$\sf{(d) \: \: {(x - r)}^{2} + {y}^{2} = {r}^{2} }$

EVALUATION

The general equation of a circle is

$\sf{{(x - \alpha )}^{2} + {(y - \beta )}^{2} = {r}^{2} } \: \: .....(1)$

$\sf{Where \: centre \: is \: ( \alpha , \beta ) \: and \: radius \: = r \: unit}$

Since the circle touches y axis, then the abscissa of the coordinates of the centre is equal to radius of the circle

$\therefore \sf{ \alpha = r}$

Equation (1) becomes

$\sf{{(x - r )}^{2} + {(y - \beta )}^{2} = {r}^{2} } \: \: .....(2)$

Now the circle touches y axis at origin

So the circle passes through the point (0,0)

From Equation (2) we get

$\sf{{(0- r )}^{2} + {(0- \beta )}^{2} = {r}^{2} } \: \:$

$\implies \sf{{ r }^{2} + { \beta }^{2} = {r}^{2} } \: \:$

$\implies \sf{ { \beta }^{2} = 0 } \: \:$

$\implies \sf{ { \beta } = 0 } \: \:$

Hence equation (2) becomes

$\sf{{(x - r )}^{2} + {y}^{2} = {r}^{2} }$

Hence the correct option is

$\sf{(d) \: \: {(x - r)}^{2} + {y}^{2} = {r}^{2} }$

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