Let the lines (2 – i) z = (2 + i) <math> <mrow> <mover accent="true"> <mi>z</mi> <mo>¯</mo> </mover> <mtext> </mtext> <mtext> </mtext> <mi>a</mi> <mi>n</mi> <mi>d</mi> <mtext> </mtext> <mtext> </mtext> <mrow> <mo>(</mo> <mrow> <mn>2</mn> <mo>+</mo> <mi>i</mi> </mrow> <mo>)</mo> </mrow> <mi>z</mi> <mo>+</mo> <mrow> <mo>(</mo> <mrow> <mi>i</mi> <mo>−</mo> <mn>2</mn> </mrow> <mo>)</mo> </mrow> <mover accent="true"> <mi>z</mi> <mo>¯</mo> </mover> <mo>−</mo> <mn>4</mn> <mi>i</mi> <mo>=</mo> <mn>0</mn> <mo>,</mo> </mrow> </math> (here i2 = -1) be normal to a circle C. If the line iz + <math> <mrow> <mover accent="true"> <mi>z</mi> <mo>¯</mo> </mover> <mo>+</mo> <mn>1</mn> <mo>+</mo> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> </math> &nbsp;is tangent to this circle C, then its radius is:

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Ncert Solutions Maths class 12th Maths Class 12th Complex Numbers and Quadratic Equations

Let the lines (2 – i) z = (2 + i) $\bar{z} a n d (2 + i) z + (i - 2) \bar{z} - 4 i = 0,$ (here i² = -1) be normal to a circle C. If the line iz + $\bar{z} + 1 + i = 0$ is tangent to this circle C, then its radius is:

Option 1 -

$\frac{3}{\sqrt[]{2}}$

Option 2 -

$\frac{3}{2 \sqrt[]{2}}$

Option 3 -

$\frac{1}{2 \sqrt[]{2}}$

Option 4 -

$3 \sqrt[]{2}$

2 Views | Posted 2 weeks ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

2 weeks ago

Correct Option - 2

Detailed Solution:

(2 – i) z = (2 + i) $\bar{z}$ , put z = x + iy

$y = \frac{x}{2} . . . . . . . . (i)$

(ii) $(2 + i) z + (i - 2) \bar{z} - 4 i = 0$

x + 2y = 2

(iii) $i z + \bar{z} + 1 + i = 0$

Equation of tangent x – y + 1 = 0

Solving (i) and (ii)

$x = 1 . y = \frac{1}{2} \Rightarrow c e n t r e (1, \frac{1}{2})$

Perpendicular distance of point $(1, \frac{1}{2})$ $\frac{}{}$ from x – y + 1 = 0 is p = r $\Rightarrow | \frac{1 - \frac{1}{2} + 1}{\sqrt[]{2}} | = r$ $\frac{\frac{}{}}{}$

$r = \frac{3}{2 \sqrt[]{2}}$

(2 – i) z = (2 + i) <math> <mrow> <mover accent="true"> <mi>z</mi> <mo>¯</mo> </mover> </mrow> </math> <math><mrow><mover accent="true"></mover></mrow></math> , put z = x + iy <math> <mrow> <mi>y</mi> <mo>=</mo> <mfrac> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mrow> <mo> (</mo> <mrow> <mi>i</mi> </mrow> <mo>)</mo> </mrow> </mrow> </math> (ii) <math> <mrow> <mrow> <mo> (</mo> <mrow> <mn>2</mn> <mo>+</mo> <mi>i</mi> </mrow> <mo>)</mo> </mrow> <mi>z</mi> <mo>+</mo> <mrow> <mo> (</mo> <mrow> <mi>i</mi> <mo>−</mo> <mn>2</mn> </mrow> <mo>)</mo> </mrow> <mover accent="true"> <mi>z</mi> <mo>¯</mo> </mover> <mo>−</mo> <mn>4</mn> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> </math> <math><mrow><mrow><mrow></mrow></mrow><mrow><mrow></mrow></mrow><mover accent="true"></mover></mrow></math>x + 2y = 2 (iii) <math> <mrow> <mi>i</mi> <mi>z</mi> <mo>+</mo> <mover accent="true"> <mi>z</mi> <mo>¯</mo> </mover> <mo>+</mo> <mn>1</mn> <mo>+</mo> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> </math> <math><mrow><mover accent="true"></mover></mrow></math>Equation of tangent x – y + 1 = 0Solving (i) and (ii) <math> <mrow> <mi>x</mi> <mo>=</mo> <mn>1</mn> <mo>.</mo> <mi>y</mi> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>⇒</mo> <mi>c</mi> <mi>e</mi> <mi>n</mi> <mi>t</mi> <mi>r</mi> <mi>e</mi> <mrow> <mo> (</mo> <mrow> <mn>1</mn> <mo>, </mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> </math> Perpendicular distance of point <math> <mrow> <mrow> <mo> (</mo> <mrow> <mn>1</mn> <mo>, </mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> </math> <math><mrow><mrow><mrow><mfrac><mrow></mrow><mrow></mrow></mfrac></mrow></mrow></mrow></math> from x – y + 1 = 0 is p = r <math> <mrow> <mo>⇒</mo> <mrow> <mo>|</mo> <mrow> <mfrac> <mrow> <mn>1</mn> <mo>−</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>+</mo> <mn>1</mn> </mrow> <mrow> <mroot> <mrow> <mn>2</mn> </mrow> <mrow></mrow> </mroot> </mrow> </mfrac> </mrow> <mo>|</mo> </mrow> <mo>=</mo> <mi>r</mi> </mrow> </math> <math><mrow><mrow><mrow><mfrac><mrow><mfrac><mrow></mrow><mrow></mrow></mfrac></mrow><mrow><mroot><mrow></mrow></mroot></mrow></mfrac></mrow></mrow></mrow></math> <math> <mrow> <mi>r</mi> <mo>=</mo> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>2</mn> <mroot> <mrow> <mn>2</mn> </mrow> <mrow></mrow> </mroot> </mrow> </mfrac> </mrow> </math>

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Let the lines (2 – i) z = (2 + i) $\bar{z} a n d (2 + i) z + (i - 2) \bar{z} - 4 i = 0,$ (here i² = -1) be normal to a circle C. If the line iz + $\bar{z} + 1 + i = 0$ is tangent to this circle C, then its radius is:

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Let the lines (2 – i) z = (2 + i) z ¯ a n d ( 2 + i ) z + ( i − 2 ) z ¯ − 4 i = 0 , (here i2 = -1) be normal to a circle C. If the line iz + z ¯ + 1 + i = 0 is tangent to this circle C, then its radius is:

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Let the lines (2 – i) z = (2 + i) $\bar{z} a n d (2 + i) z + (i - 2) \bar{z} - 4 i = 0,$ (here i² = -1) be normal to a circle C. If the line iz + $\bar{z} + 1 + i = 0$ is tangent to this circle C, then its radius is: