QBManager

Conic Section

119699 The equation of circle with centre $(2, - 3)$ and touching $x$ -axis is

1

x^{2} + y^{2} - 4 x - 6 y + 4 = 0

2

x^{2} + y^{2} - 4 x - 6 y - 8 = 0

3

x^{2} + y^{2} - 4 x + 6 y + 4 = 0

4

x^{2} + y^{2} + 4 x - 6 y + 8 = 0

Explanation:

C Given centre (2,-3)
Equation of circle,
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
and, centre $\to (- g, - f)$
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
$x^{2} + y^{2} - 4 x + 6 y + c = 0$
touching $x$ axis then $y = 0$
$4 - 8 + c = 0$
$c = 4$
Putting the value $c$ in equation
$x^{2} + y^{2} - 4 x + 6 y + 4 = 0$

AP EAMCET-07.07.2022

Conic Section

119701 For all real values of $k$ , the polar of the point $(2 k, k - 4)$ with respect to $x^{2} + y^{2} - 4 x - 6 y + 1 =$ 0 passes through the point

1

(1, 1)

2

(1, - 1)

3

(- 3, 1)

4

(3, 1)

Explanation:

D Given equation is -
$x^{2} + y^{2} - 4 x - 6 y + 1 = 0$
Equation of polar is $T = 0$
${xx}_{1} + {yy}_{1} - 4 (\frac{x + x_{1}}{2}) - 6 (\frac{y + y_{1}}{2}) + 1 = 0$
$x (2 k) + y (k - 4) - 2 (x + 2 k) - 3 (y + k - 4) + 1 = 0$
$2 kx + yk - 4 y - 2 x - 4 k - 3 y - 3 k + 12 + 1 = 0$
$k (2 x + y - 7) - (2 x + 7 y - 13) = 0$
The equation of polar will be independent of $k$ if its coefficient is zero.
$2 x + y - 7 = 0$
$2 x + 7 y - 13 = 0$
$- +$
$- 6 y + 6 = 0$
$y = 1$
$x = 3$ So polar passes through the fix point $(3, 1)$

TS EAMCET-2016

Conic Section

119702 The value of a, such that the power of the point $(1, 6)$ with respect to the circle $x^{2} + y^{2} + 4 x - 6 y$ $- a = 0$ , is $- 16$ is,

1 7

2 11

3 13

4 21

Explanation:

D Given circle equation,
$x^{2} + y^{2} + 4 x - 6 y - a = 0$
We know that power of point $p (1, 6)$ with respect the
circle
$x^{2} + y^{2} + 4 x - 6 y - a = 0$ is -16
Then,
$(1)^{2} + (6)^{2} + 4 \times 1 - 6 \times 6 - a = - 16$
$1 + 36 + 4 - 36 - a = - 16$
$a = 16 + 5$
$a = 21$

TS EAMCET-2015

Conic Section

119703 The radius of the circle passing through the points $(- 1, 1), (2, - 1)$ and $(1, 0)$ is

1 5

2

\frac{\sqrt{130}}{2}

3 6

4

\frac{\sqrt{145}}{2}

Explanation:

B Given point $(- 1, 1) (2, - 1), (1, 0)$
General equation of circle is -
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
At point $(- 1, 1)$
$(1)^{2} + (1)^{2} + 2 g (- 1) + 2 f (1) + c = 0$
$- 2 g + 2 f + c + 2 = 0$
At point $(2, - 1)$
$(2)^{2} + (- 1)^{2} + 2 g (2) + 2 f (- 1) + c = 0$
$4 g - 2 f + c + 5 = 0$
By (i) and (ii), we get -
$2 g + 2 c + 7 = 0$
At point $(1, 0)$
$(1)^{2} + 0 + 2 g + 0 + c = 0$
$2 g + c + 1 = 0$
By solving equation (iii) and (iv) we get -
$c = - 6, g = 5 / 2$
By putting the value of $c & g$ in equation (i) we get -
$f = 9 / 2$
Now, Radius of the circle is -
$r = \sqrt{g^{2} + f^{2} - c}$
$= \sqrt{{(\frac{5}{2})}^{2} + {(\frac{9}{2})}^{2} + 6} = \frac{\sqrt{130}}{2}$

TS EAMCET-19.07.2022

Conic Section

119704 If the circle $x^{2} + y^{2} - 6 x + 2 y = 28$ cuts off a chord of length $λ$ units on the line $2 x - 5 y + 18$ $= 0$ , then the value of $λ$ is

1 3

2 6

3 12

4 9

Explanation:

B Given circle equation
original image
$x^{2} + y^{2} - 6 x + 2 y = 28$
by comparing general equation of circle we get
$g = - 3 f = 1, c = - 28$
$Radius (r) = \sqrt{g^{2} + f^{2} - c} = \sqrt{(- 3)^{2} + (1)^{2} - (- 28)} = \sqrt{38}$
$CP = \frac{2 \times 3 - 5 \times (- 1) + 18}{\sqrt{(2)^{2} + (- 5)^{2}}}$
$CP = \frac{29}{\sqrt{29}} = \sqrt{29}$
$From Δ CPA$
${AC}^{2} = {CP}^{2} + (AP)^{2}$
$(\sqrt{38})^{2} = (\sqrt{29})^{2} + (AP)^{2}$
$38 = 29 + (AP)^{2}$
$(AP)^{2} = 38 - 29 = 9$
$AP = 3$ Lenth of chord of the circle $= 2 \times AP = 2 \times 3 = 6$

TS EAMCET-09.09.2020

Conic Section

119699 The equation of circle with centre $(2, - 3)$ and touching $x$ -axis is

1

x^{2} + y^{2} - 4 x - 6 y + 4 = 0

2

x^{2} + y^{2} - 4 x - 6 y - 8 = 0

3

x^{2} + y^{2} - 4 x + 6 y + 4 = 0

4

x^{2} + y^{2} + 4 x - 6 y + 8 = 0

Explanation:

C Given centre (2,-3)
Equation of circle,
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
and, centre $\to (- g, - f)$
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
$x^{2} + y^{2} - 4 x + 6 y + c = 0$
touching $x$ axis then $y = 0$
$4 - 8 + c = 0$
$c = 4$
Putting the value $c$ in equation
$x^{2} + y^{2} - 4 x + 6 y + 4 = 0$

AP EAMCET-07.07.2022

Conic Section

119701 For all real values of $k$ , the polar of the point $(2 k, k - 4)$ with respect to $x^{2} + y^{2} - 4 x - 6 y + 1 =$ 0 passes through the point

1

(1, 1)

2

(1, - 1)

3

(- 3, 1)

4

(3, 1)

Explanation:

D Given equation is -
$x^{2} + y^{2} - 4 x - 6 y + 1 = 0$
Equation of polar is $T = 0$
${xx}_{1} + {yy}_{1} - 4 (\frac{x + x_{1}}{2}) - 6 (\frac{y + y_{1}}{2}) + 1 = 0$
$x (2 k) + y (k - 4) - 2 (x + 2 k) - 3 (y + k - 4) + 1 = 0$
$2 kx + yk - 4 y - 2 x - 4 k - 3 y - 3 k + 12 + 1 = 0$
$k (2 x + y - 7) - (2 x + 7 y - 13) = 0$
The equation of polar will be independent of $k$ if its coefficient is zero.
$2 x + y - 7 = 0$
$2 x + 7 y - 13 = 0$
$- +$
$- 6 y + 6 = 0$
$y = 1$
$x = 3$ So polar passes through the fix point $(3, 1)$

TS EAMCET-2016

Conic Section

119702 The value of a, such that the power of the point $(1, 6)$ with respect to the circle $x^{2} + y^{2} + 4 x - 6 y$ $- a = 0$ , is $- 16$ is,

1 7

2 11

3 13

4 21

Explanation:

D Given circle equation,
$x^{2} + y^{2} + 4 x - 6 y - a = 0$
We know that power of point $p (1, 6)$ with respect the
circle
$x^{2} + y^{2} + 4 x - 6 y - a = 0$ is -16
Then,
$(1)^{2} + (6)^{2} + 4 \times 1 - 6 \times 6 - a = - 16$
$1 + 36 + 4 - 36 - a = - 16$
$a = 16 + 5$
$a = 21$

TS EAMCET-2015

Conic Section

119703 The radius of the circle passing through the points $(- 1, 1), (2, - 1)$ and $(1, 0)$ is

1 5

2

\frac{\sqrt{130}}{2}

3 6

4

\frac{\sqrt{145}}{2}

Explanation:

B Given point $(- 1, 1) (2, - 1), (1, 0)$
General equation of circle is -
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
At point $(- 1, 1)$
$(1)^{2} + (1)^{2} + 2 g (- 1) + 2 f (1) + c = 0$
$- 2 g + 2 f + c + 2 = 0$
At point $(2, - 1)$
$(2)^{2} + (- 1)^{2} + 2 g (2) + 2 f (- 1) + c = 0$
$4 g - 2 f + c + 5 = 0$
By (i) and (ii), we get -
$2 g + 2 c + 7 = 0$
At point $(1, 0)$
$(1)^{2} + 0 + 2 g + 0 + c = 0$
$2 g + c + 1 = 0$
By solving equation (iii) and (iv) we get -
$c = - 6, g = 5 / 2$
By putting the value of $c & g$ in equation (i) we get -
$f = 9 / 2$
Now, Radius of the circle is -
$r = \sqrt{g^{2} + f^{2} - c}$
$= \sqrt{{(\frac{5}{2})}^{2} + {(\frac{9}{2})}^{2} + 6} = \frac{\sqrt{130}}{2}$

TS EAMCET-19.07.2022

Conic Section

119704 If the circle $x^{2} + y^{2} - 6 x + 2 y = 28$ cuts off a chord of length $λ$ units on the line $2 x - 5 y + 18$ $= 0$ , then the value of $λ$ is

1 3

2 6

3 12

4 9

Explanation:

B Given circle equation
original image
$x^{2} + y^{2} - 6 x + 2 y = 28$
by comparing general equation of circle we get
$g = - 3 f = 1, c = - 28$
$Radius (r) = \sqrt{g^{2} + f^{2} - c} = \sqrt{(- 3)^{2} + (1)^{2} - (- 28)} = \sqrt{38}$
$CP = \frac{2 \times 3 - 5 \times (- 1) + 18}{\sqrt{(2)^{2} + (- 5)^{2}}}$
$CP = \frac{29}{\sqrt{29}} = \sqrt{29}$
$From Δ CPA$
${AC}^{2} = {CP}^{2} + (AP)^{2}$
$(\sqrt{38})^{2} = (\sqrt{29})^{2} + (AP)^{2}$
$38 = 29 + (AP)^{2}$
$(AP)^{2} = 38 - 29 = 9$
$AP = 3$ Lenth of chord of the circle $= 2 \times AP = 2 \times 3 = 6$

TS EAMCET-09.09.2020

Conic Section

119699 The equation of circle with centre $(2, - 3)$ and touching $x$ -axis is

1

x^{2} + y^{2} - 4 x - 6 y + 4 = 0

2

x^{2} + y^{2} - 4 x - 6 y - 8 = 0

3

x^{2} + y^{2} - 4 x + 6 y + 4 = 0

4

x^{2} + y^{2} + 4 x - 6 y + 8 = 0

Explanation:

C Given centre (2,-3)
Equation of circle,
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
and, centre $\to (- g, - f)$
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
$x^{2} + y^{2} - 4 x + 6 y + c = 0$
touching $x$ axis then $y = 0$
$4 - 8 + c = 0$
$c = 4$
Putting the value $c$ in equation
$x^{2} + y^{2} - 4 x + 6 y + 4 = 0$

AP EAMCET-07.07.2022

Conic Section

119701 For all real values of $k$ , the polar of the point $(2 k, k - 4)$ with respect to $x^{2} + y^{2} - 4 x - 6 y + 1 =$ 0 passes through the point

1

(1, 1)

2

(1, - 1)

3

(- 3, 1)

4

(3, 1)

Explanation:

D Given equation is -
$x^{2} + y^{2} - 4 x - 6 y + 1 = 0$
Equation of polar is $T = 0$
${xx}_{1} + {yy}_{1} - 4 (\frac{x + x_{1}}{2}) - 6 (\frac{y + y_{1}}{2}) + 1 = 0$
$x (2 k) + y (k - 4) - 2 (x + 2 k) - 3 (y + k - 4) + 1 = 0$
$2 kx + yk - 4 y - 2 x - 4 k - 3 y - 3 k + 12 + 1 = 0$
$k (2 x + y - 7) - (2 x + 7 y - 13) = 0$
The equation of polar will be independent of $k$ if its coefficient is zero.
$2 x + y - 7 = 0$
$2 x + 7 y - 13 = 0$
$- +$
$- 6 y + 6 = 0$
$y = 1$
$x = 3$ So polar passes through the fix point $(3, 1)$

TS EAMCET-2016

Conic Section

119702 The value of a, such that the power of the point $(1, 6)$ with respect to the circle $x^{2} + y^{2} + 4 x - 6 y$ $- a = 0$ , is $- 16$ is,

1 7

2 11

3 13

4 21

Explanation:

D Given circle equation,
$x^{2} + y^{2} + 4 x - 6 y - a = 0$
We know that power of point $p (1, 6)$ with respect the
circle
$x^{2} + y^{2} + 4 x - 6 y - a = 0$ is -16
Then,
$(1)^{2} + (6)^{2} + 4 \times 1 - 6 \times 6 - a = - 16$
$1 + 36 + 4 - 36 - a = - 16$
$a = 16 + 5$
$a = 21$

TS EAMCET-2015

Conic Section

119703 The radius of the circle passing through the points $(- 1, 1), (2, - 1)$ and $(1, 0)$ is

1 5

2

\frac{\sqrt{130}}{2}

3 6

4

\frac{\sqrt{145}}{2}

Explanation:

B Given point $(- 1, 1) (2, - 1), (1, 0)$
General equation of circle is -
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
At point $(- 1, 1)$
$(1)^{2} + (1)^{2} + 2 g (- 1) + 2 f (1) + c = 0$
$- 2 g + 2 f + c + 2 = 0$
At point $(2, - 1)$
$(2)^{2} + (- 1)^{2} + 2 g (2) + 2 f (- 1) + c = 0$
$4 g - 2 f + c + 5 = 0$
By (i) and (ii), we get -
$2 g + 2 c + 7 = 0$
At point $(1, 0)$
$(1)^{2} + 0 + 2 g + 0 + c = 0$
$2 g + c + 1 = 0$
By solving equation (iii) and (iv) we get -
$c = - 6, g = 5 / 2$
By putting the value of $c & g$ in equation (i) we get -
$f = 9 / 2$
Now, Radius of the circle is -
$r = \sqrt{g^{2} + f^{2} - c}$
$= \sqrt{{(\frac{5}{2})}^{2} + {(\frac{9}{2})}^{2} + 6} = \frac{\sqrt{130}}{2}$

TS EAMCET-19.07.2022

Conic Section

119704 If the circle $x^{2} + y^{2} - 6 x + 2 y = 28$ cuts off a chord of length $λ$ units on the line $2 x - 5 y + 18$ $= 0$ , then the value of $λ$ is

1 3

2 6

3 12

4 9

Explanation:

B Given circle equation
original image
$x^{2} + y^{2} - 6 x + 2 y = 28$
by comparing general equation of circle we get
$g = - 3 f = 1, c = - 28$
$Radius (r) = \sqrt{g^{2} + f^{2} - c} = \sqrt{(- 3)^{2} + (1)^{2} - (- 28)} = \sqrt{38}$
$CP = \frac{2 \times 3 - 5 \times (- 1) + 18}{\sqrt{(2)^{2} + (- 5)^{2}}}$
$CP = \frac{29}{\sqrt{29}} = \sqrt{29}$
$From Δ CPA$
${AC}^{2} = {CP}^{2} + (AP)^{2}$
$(\sqrt{38})^{2} = (\sqrt{29})^{2} + (AP)^{2}$
$38 = 29 + (AP)^{2}$
$(AP)^{2} = 38 - 29 = 9$
$AP = 3$ Lenth of chord of the circle $= 2 \times AP = 2 \times 3 = 6$

TS EAMCET-09.09.2020

Conic Section

119699 The equation of circle with centre $(2, - 3)$ and touching $x$ -axis is

1

x^{2} + y^{2} - 4 x - 6 y + 4 = 0

2

x^{2} + y^{2} - 4 x - 6 y - 8 = 0

3

x^{2} + y^{2} - 4 x + 6 y + 4 = 0

4

x^{2} + y^{2} + 4 x - 6 y + 8 = 0

Explanation:

C Given centre (2,-3)
Equation of circle,
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
and, centre $\to (- g, - f)$
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
$x^{2} + y^{2} - 4 x + 6 y + c = 0$
touching $x$ axis then $y = 0$
$4 - 8 + c = 0$
$c = 4$
Putting the value $c$ in equation
$x^{2} + y^{2} - 4 x + 6 y + 4 = 0$

AP EAMCET-07.07.2022

Conic Section

119701 For all real values of $k$ , the polar of the point $(2 k, k - 4)$ with respect to $x^{2} + y^{2} - 4 x - 6 y + 1 =$ 0 passes through the point

1

(1, 1)

2

(1, - 1)

3

(- 3, 1)

4

(3, 1)

Explanation:

D Given equation is -
$x^{2} + y^{2} - 4 x - 6 y + 1 = 0$
Equation of polar is $T = 0$
${xx}_{1} + {yy}_{1} - 4 (\frac{x + x_{1}}{2}) - 6 (\frac{y + y_{1}}{2}) + 1 = 0$
$x (2 k) + y (k - 4) - 2 (x + 2 k) - 3 (y + k - 4) + 1 = 0$
$2 kx + yk - 4 y - 2 x - 4 k - 3 y - 3 k + 12 + 1 = 0$
$k (2 x + y - 7) - (2 x + 7 y - 13) = 0$
The equation of polar will be independent of $k$ if its coefficient is zero.
$2 x + y - 7 = 0$
$2 x + 7 y - 13 = 0$
$- +$
$- 6 y + 6 = 0$
$y = 1$
$x = 3$ So polar passes through the fix point $(3, 1)$

TS EAMCET-2016

Conic Section

119702 The value of a, such that the power of the point $(1, 6)$ with respect to the circle $x^{2} + y^{2} + 4 x - 6 y$ $- a = 0$ , is $- 16$ is,

1 7

2 11

3 13

4 21

Explanation:

D Given circle equation,
$x^{2} + y^{2} + 4 x - 6 y - a = 0$
We know that power of point $p (1, 6)$ with respect the
circle
$x^{2} + y^{2} + 4 x - 6 y - a = 0$ is -16
Then,
$(1)^{2} + (6)^{2} + 4 \times 1 - 6 \times 6 - a = - 16$
$1 + 36 + 4 - 36 - a = - 16$
$a = 16 + 5$
$a = 21$

TS EAMCET-2015

Conic Section

119703 The radius of the circle passing through the points $(- 1, 1), (2, - 1)$ and $(1, 0)$ is

1 5

2

\frac{\sqrt{130}}{2}

3 6

4

\frac{\sqrt{145}}{2}

Explanation:

B Given point $(- 1, 1) (2, - 1), (1, 0)$
General equation of circle is -
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
At point $(- 1, 1)$
$(1)^{2} + (1)^{2} + 2 g (- 1) + 2 f (1) + c = 0$
$- 2 g + 2 f + c + 2 = 0$
At point $(2, - 1)$
$(2)^{2} + (- 1)^{2} + 2 g (2) + 2 f (- 1) + c = 0$
$4 g - 2 f + c + 5 = 0$
By (i) and (ii), we get -
$2 g + 2 c + 7 = 0$
At point $(1, 0)$
$(1)^{2} + 0 + 2 g + 0 + c = 0$
$2 g + c + 1 = 0$
By solving equation (iii) and (iv) we get -
$c = - 6, g = 5 / 2$
By putting the value of $c & g$ in equation (i) we get -
$f = 9 / 2$
Now, Radius of the circle is -
$r = \sqrt{g^{2} + f^{2} - c}$
$= \sqrt{{(\frac{5}{2})}^{2} + {(\frac{9}{2})}^{2} + 6} = \frac{\sqrt{130}}{2}$

TS EAMCET-19.07.2022

Conic Section

119704 If the circle $x^{2} + y^{2} - 6 x + 2 y = 28$ cuts off a chord of length $λ$ units on the line $2 x - 5 y + 18$ $= 0$ , then the value of $λ$ is

1 3

2 6

3 12

4 9

Explanation:

B Given circle equation
original image
$x^{2} + y^{2} - 6 x + 2 y = 28$
by comparing general equation of circle we get
$g = - 3 f = 1, c = - 28$
$Radius (r) = \sqrt{g^{2} + f^{2} - c} = \sqrt{(- 3)^{2} + (1)^{2} - (- 28)} = \sqrt{38}$
$CP = \frac{2 \times 3 - 5 \times (- 1) + 18}{\sqrt{(2)^{2} + (- 5)^{2}}}$
$CP = \frac{29}{\sqrt{29}} = \sqrt{29}$
$From Δ CPA$
${AC}^{2} = {CP}^{2} + (AP)^{2}$
$(\sqrt{38})^{2} = (\sqrt{29})^{2} + (AP)^{2}$
$38 = 29 + (AP)^{2}$
$(AP)^{2} = 38 - 29 = 9$
$AP = 3$ Lenth of chord of the circle $= 2 \times AP = 2 \times 3 = 6$

TS EAMCET-09.09.2020

Conic Section

119699 The equation of circle with centre $(2, - 3)$ and touching $x$ -axis is

1

x^{2} + y^{2} - 4 x - 6 y + 4 = 0

2

x^{2} + y^{2} - 4 x - 6 y - 8 = 0

3

x^{2} + y^{2} - 4 x + 6 y + 4 = 0

4

x^{2} + y^{2} + 4 x - 6 y + 8 = 0

Explanation:

C Given centre (2,-3)
Equation of circle,
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
and, centre $\to (- g, - f)$
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
$x^{2} + y^{2} - 4 x + 6 y + c = 0$
touching $x$ axis then $y = 0$
$4 - 8 + c = 0$
$c = 4$
Putting the value $c$ in equation
$x^{2} + y^{2} - 4 x + 6 y + 4 = 0$

AP EAMCET-07.07.2022

Conic Section

119701 For all real values of $k$ , the polar of the point $(2 k, k - 4)$ with respect to $x^{2} + y^{2} - 4 x - 6 y + 1 =$ 0 passes through the point

1

(1, 1)

2

(1, - 1)

3

(- 3, 1)

4

(3, 1)

Explanation:

D Given equation is -
$x^{2} + y^{2} - 4 x - 6 y + 1 = 0$
Equation of polar is $T = 0$
${xx}_{1} + {yy}_{1} - 4 (\frac{x + x_{1}}{2}) - 6 (\frac{y + y_{1}}{2}) + 1 = 0$
$x (2 k) + y (k - 4) - 2 (x + 2 k) - 3 (y + k - 4) + 1 = 0$
$2 kx + yk - 4 y - 2 x - 4 k - 3 y - 3 k + 12 + 1 = 0$
$k (2 x + y - 7) - (2 x + 7 y - 13) = 0$
The equation of polar will be independent of $k$ if its coefficient is zero.
$2 x + y - 7 = 0$
$2 x + 7 y - 13 = 0$
$- +$
$- 6 y + 6 = 0$
$y = 1$
$x = 3$ So polar passes through the fix point $(3, 1)$

TS EAMCET-2016

Conic Section

119702 The value of a, such that the power of the point $(1, 6)$ with respect to the circle $x^{2} + y^{2} + 4 x - 6 y$ $- a = 0$ , is $- 16$ is,

1 7

2 11

3 13

4 21

Explanation:

D Given circle equation,
$x^{2} + y^{2} + 4 x - 6 y - a = 0$
We know that power of point $p (1, 6)$ with respect the
circle
$x^{2} + y^{2} + 4 x - 6 y - a = 0$ is -16
Then,
$(1)^{2} + (6)^{2} + 4 \times 1 - 6 \times 6 - a = - 16$
$1 + 36 + 4 - 36 - a = - 16$
$a = 16 + 5$
$a = 21$

TS EAMCET-2015

Conic Section

119703 The radius of the circle passing through the points $(- 1, 1), (2, - 1)$ and $(1, 0)$ is

1 5

2

\frac{\sqrt{130}}{2}

3 6

4

\frac{\sqrt{145}}{2}

Explanation:

B Given point $(- 1, 1) (2, - 1), (1, 0)$
General equation of circle is -
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
At point $(- 1, 1)$
$(1)^{2} + (1)^{2} + 2 g (- 1) + 2 f (1) + c = 0$
$- 2 g + 2 f + c + 2 = 0$
At point $(2, - 1)$
$(2)^{2} + (- 1)^{2} + 2 g (2) + 2 f (- 1) + c = 0$
$4 g - 2 f + c + 5 = 0$
By (i) and (ii), we get -
$2 g + 2 c + 7 = 0$
At point $(1, 0)$
$(1)^{2} + 0 + 2 g + 0 + c = 0$
$2 g + c + 1 = 0$
By solving equation (iii) and (iv) we get -
$c = - 6, g = 5 / 2$
By putting the value of $c & g$ in equation (i) we get -
$f = 9 / 2$
Now, Radius of the circle is -
$r = \sqrt{g^{2} + f^{2} - c}$
$= \sqrt{{(\frac{5}{2})}^{2} + {(\frac{9}{2})}^{2} + 6} = \frac{\sqrt{130}}{2}$

TS EAMCET-19.07.2022

Conic Section

119704 If the circle $x^{2} + y^{2} - 6 x + 2 y = 28$ cuts off a chord of length $λ$ units on the line $2 x - 5 y + 18$ $= 0$ , then the value of $λ$ is

1 3

2 6

3 12

4 9

Explanation:

B Given circle equation
original image
$x^{2} + y^{2} - 6 x + 2 y = 28$
by comparing general equation of circle we get
$g = - 3 f = 1, c = - 28$
$Radius (r) = \sqrt{g^{2} + f^{2} - c} = \sqrt{(- 3)^{2} + (1)^{2} - (- 28)} = \sqrt{38}$
$CP = \frac{2 \times 3 - 5 \times (- 1) + 18}{\sqrt{(2)^{2} + (- 5)^{2}}}$
$CP = \frac{29}{\sqrt{29}} = \sqrt{29}$
$From Δ CPA$
${AC}^{2} = {CP}^{2} + (AP)^{2}$
$(\sqrt{38})^{2} = (\sqrt{29})^{2} + (AP)^{2}$
$38 = 29 + (AP)^{2}$
$(AP)^{2} = 38 - 29 = 9$
$AP = 3$ Lenth of chord of the circle $= 2 \times AP = 2 \times 3 = 6$

TS EAMCET-09.09.2020