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The circle passing through the point (-1,0) and touching the Y-axis at (0, 2) also passes through the point
The circle passing through the point (-1,0) and touching the Y-axis at (0, 2) also passes through the point
Updated On: Jun 14, 2022
- $\bigg(\frac{3}{2},0\bigg)$
- $\bigg(-\frac{5}{2},2\bigg)$
- $\bigg(-\frac{3}{2},\frac{5}{2}\bigg)$
- $(-4, 0)$
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The Correct Option is D
Solution and Explanation
Equation of circle passing through a point $(x_1, y_1)$ and
touching the straight line L, is given by
$( x - x_1 )^2 + ( y - y_1)^2 + \lambdaL = 0$
$\therefore$ Equation of circle passing through (0, 2) and touching x=0
$\Rightarrow \, \, \, \, \, \, \, ( x -0)^2+ (y -2 )^2+ \lambdax = 0\hspace15mm ...(i)$
Also, it passes through $(-1 ,0 ) \Rightarrow 1 + 4 - \lambda = 0$
$\therefore$ \lambda=5
E (i) becomes,
$\hspace22mm x^2 + y^2 - 4y + 4 + 5x = 0$
$\Rightarrow\hspace15mm x^2 + y^2 - 5x - 4y+4 =0$,
For x-intercept put $y = 0 \Rightarrow x^2 + 5x + 4 = 0,$
$(x + 1) (x+4)=0$
$\therefore$ $\hspace20mm $x=-1, -4
Hence, (d) option (-4, 0).
touching the straight line L, is given by
$( x - x_1 )^2 + ( y - y_1)^2 + \lambdaL = 0$
$\therefore$ Equation of circle passing through (0, 2) and touching x=0
$\Rightarrow \, \, \, \, \, \, \, ( x -0)^2+ (y -2 )^2+ \lambdax = 0\hspace15mm ...(i)$
Also, it passes through $(-1 ,0 ) \Rightarrow 1 + 4 - \lambda = 0$
$\therefore$ \lambda=5
E (i) becomes,
$\hspace22mm x^2 + y^2 - 4y + 4 + 5x = 0$
$\Rightarrow\hspace15mm x^2 + y^2 - 5x - 4y+4 =0$,
For x-intercept put $y = 0 \Rightarrow x^2 + 5x + 4 = 0,$
$(x + 1) (x+4)=0$
$\therefore$ $\hspace20mm $x=-1, -4
Hence, (d) option (-4, 0).
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