QBManager

Conic Section

119756 If the lines $x + 2 y - 5 = 0$ and $3 x - y - 1 = 0$ denote two diameters of a circle of radius 5 units, then the equation of the circle is

1

x^{2} + y^{2} - 2 x + 4 y - 20 = 0

2

x^{2} + y^{2} - 2 x - 4 y - 20 = 0

3

x^{2} + y^{2} + 2 x - 4 y + 20 = 0

4

x^{2} + y^{2} + 2 x + 4 y + 20 = 0

Explanation:

B Equation of line are:-
$x + 2 y = 5$
$3 x - y = 1$
So,
$3 x + 6 y = 15$
$- 3 x + y = - 1$
Subtract (i) and (ii)
$\overset{―}{7 y} = 14 / 2$
$y = 2$
Put $y = 2$ in $3 x - y = 1$
$3 x - 2 = 1$
$3 x = 3$
$x = 1$
So, equation of circle is:-
$(x - 1)^{2} + (x - 2)^{2} = 5^{2}$
$x^{2} + 1 - 2 x + y^{2} + 4 - 4 y = 25$
$x^{2} + y^{2} - 2 x - 4 y - 20 = 0$

APEAPCET-20.08.2021

Conic Section

119757 Find the equation of a circle which cuts the circle $x^{2} + y^{2} - 6 x + 4 y - 3 = 0$ orthogonally while passing through $(3, 0)$ and touching the $y$ axis.

1

x^{2} + y^{2} + 6 x + 6 y + 9 = 0

2

x^{2} + y^{2} - 6 x - 6 y + 9 = 0

3

x^{2} + y^{2} - 6 x + 6 y - 9 = 0

4

x^{2} + y^{2} + 6 x - 6 y - 9 = 0

Explanation:

B Let, the equation of circle is:-
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
$Touches y-axis then f^{2} = c$
$Since (i) passes through (3, 0) then$
$9 + 6 g + c = 0$
$6 g + c = - 9$
$Given circle,$
$x^{2} + y^{2} - 6 x + 4 y - 3 = 0$
Then,
$g^{'} = - 3$
$f = 2$
$c^{'} = - 3$
Since (i) and (iii) are orthogonal then
$2 g^{'} + 2 f f = c + c^{'}$
$2 g (- 3) + 2 f (2) = c - 3$
$- 6 g + 4 f - c = - 3$
Then equation (iv) and (ii) add
$4 f = - 12$
$f = - 3$
Now,
$c = 9$
From equation (ii)
$6 g + 9 = - 9$
$6 g = - 18$
$g = 3$
$From (i) circle is$
$x^{2} + y^{2} - 6 x - 6 y + 9 = 0$ From (i) circle is

APEAPCET-23.08.2021

Conic Section

119758 Suppose a circle passes through $(2, 2)$ and $(9, 9)$ and touches the $x$ -axis at $P$ . If $O$ is the origin. then $O P$ is equal to

1 4

2 5

3 6

4 9

Explanation:

C
$r = \sqrt{(h - h)^{2} + (k - 0)^{2}}$
$r = \sqrt{k^{2}}$
$r^{2} = k^{2}$
$\Rightarrow CA = CB$
$\Rightarrow (CA)^{2} = (CB)^{2}$
$(h - 2)^{2} + (k - 2)^{2} = (h - 9)^{2} + (k - 9)^{2}$
$h^{2} + k^{2} + 4 + 4 - 4 h - 4 k = h^{2} + k^{2} + 81 + 81 - 18 h - 18 k$
$14 h + 14 k = 81 + 81 - 4 - 4$
$14 (h + k) = 154$
$(h + k) = 11$
$CA = r$
$(CA)^{2} = (r)^{2}$
$(h - 2)^{2} + (k - 2)^{2} = r^{2}$
$h^{2} + 4 - 4 h + k^{2} + 4 - 4 k = k^{2} (∵ r^{2} = k^{2})$
$h^{2} + 8 - 4 h - 4 (11 - h) = 0$
$h^{2} + 8 - 4 h - 44 + 4 h = 0$
$h^{2} = 36$
$h = \pm 6$
$P \equiv (\pm 6, 0), O \equiv (0, 0)$
$OP = 6 units$
original image

APEAPCET- 23.08.2021

Conic Section

119759 Find the equation of the circle which passes through origin and cuts off the intercepts - 2 and 3 over the $x$ and $y$ axes respectively.

1

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

2

2 (x^{2} + y^{2}) + 2 x - 3 y = 0

3

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

4

x^{2} + y^{2} + 2 x - 3 y = 0

Explanation:

D Given, $x intercept = - 2, y intercept = 3$
Let the circle be
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
$(0, 0), (- 2, 0), (0, 3)$
$(0, 0) \to c = 0$
$(- 2, 0) \to 4 - 4 g = 0$
$g = 1$
$(0, 3) \to 9 + 6 f = 0$
$6 f = - 9$
$f = \frac{- 9}{6} = - 3 / 2$
Required equation of circle
$x^{2} + y^{2} + 2 x - 3 y = 0$

AP EAPCET-25.08.2021

Conic Section

119760 The equation of the circle passing through $(0$ , $0)$ and which makes intercepts $a$ and $b$ on the coordinate axes is

1

x^{2} + y^{2} + a x + b y = 0

2

x^{2} + y^{2} - a x - b y = 0

3

x^{2} + y^{2} - a x + b y = 0

4

x^{2} + y^{2} - a x - b x = 0

Explanation:

C $: Centre = (\frac{a}{2}, \frac{b}{2})$
$Radius = \frac{\sqrt{a^{2} + b^{2}}}{2}$
$Equation of circle$
original image
${(x - \frac{a}{2})}^{2} + {(x - \frac{b}{2})}^{2} = \frac{a^{2} + b^{2}}{2}$
$4 x^{2} - 4 a x + a^{2} + 4 y^{2} - 4 b y + b^{2} = a^{2} + b^{2}$
$4 x^{2} + 4 y^{2} - 4 a x - 4 b y = 0$
$4 (x^{2} + y^{2} - a x - b y) = 0$
$x^{2} + y^{2} - a x - b y = 0$

AP EAMCET-17.09.2020

Conic Section

119756 If the lines $x + 2 y - 5 = 0$ and $3 x - y - 1 = 0$ denote two diameters of a circle of radius 5 units, then the equation of the circle is

1

x^{2} + y^{2} - 2 x + 4 y - 20 = 0

2

x^{2} + y^{2} - 2 x - 4 y - 20 = 0

3

x^{2} + y^{2} + 2 x - 4 y + 20 = 0

4

x^{2} + y^{2} + 2 x + 4 y + 20 = 0

Explanation:

B Equation of line are:-
$x + 2 y = 5$
$3 x - y = 1$
So,
$3 x + 6 y = 15$
$- 3 x + y = - 1$
Subtract (i) and (ii)
$\overset{―}{7 y} = 14 / 2$
$y = 2$
Put $y = 2$ in $3 x - y = 1$
$3 x - 2 = 1$
$3 x = 3$
$x = 1$
So, equation of circle is:-
$(x - 1)^{2} + (x - 2)^{2} = 5^{2}$
$x^{2} + 1 - 2 x + y^{2} + 4 - 4 y = 25$
$x^{2} + y^{2} - 2 x - 4 y - 20 = 0$

APEAPCET-20.08.2021

Conic Section

119757 Find the equation of a circle which cuts the circle $x^{2} + y^{2} - 6 x + 4 y - 3 = 0$ orthogonally while passing through $(3, 0)$ and touching the $y$ axis.

1

x^{2} + y^{2} + 6 x + 6 y + 9 = 0

2

x^{2} + y^{2} - 6 x - 6 y + 9 = 0

3

x^{2} + y^{2} - 6 x + 6 y - 9 = 0

4

x^{2} + y^{2} + 6 x - 6 y - 9 = 0

Explanation:

B Let, the equation of circle is:-
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
$Touches y-axis then f^{2} = c$
$Since (i) passes through (3, 0) then$
$9 + 6 g + c = 0$
$6 g + c = - 9$
$Given circle,$
$x^{2} + y^{2} - 6 x + 4 y - 3 = 0$
Then,
$g^{'} = - 3$
$f = 2$
$c^{'} = - 3$
Since (i) and (iii) are orthogonal then
$2 g^{'} + 2 f f = c + c^{'}$
$2 g (- 3) + 2 f (2) = c - 3$
$- 6 g + 4 f - c = - 3$
Then equation (iv) and (ii) add
$4 f = - 12$
$f = - 3$
Now,
$c = 9$
From equation (ii)
$6 g + 9 = - 9$
$6 g = - 18$
$g = 3$
$From (i) circle is$
$x^{2} + y^{2} - 6 x - 6 y + 9 = 0$ From (i) circle is

APEAPCET-23.08.2021

Conic Section

119758 Suppose a circle passes through $(2, 2)$ and $(9, 9)$ and touches the $x$ -axis at $P$ . If $O$ is the origin. then $O P$ is equal to

1 4

2 5

3 6

4 9

Explanation:

C
$r = \sqrt{(h - h)^{2} + (k - 0)^{2}}$
$r = \sqrt{k^{2}}$
$r^{2} = k^{2}$
$\Rightarrow CA = CB$
$\Rightarrow (CA)^{2} = (CB)^{2}$
$(h - 2)^{2} + (k - 2)^{2} = (h - 9)^{2} + (k - 9)^{2}$
$h^{2} + k^{2} + 4 + 4 - 4 h - 4 k = h^{2} + k^{2} + 81 + 81 - 18 h - 18 k$
$14 h + 14 k = 81 + 81 - 4 - 4$
$14 (h + k) = 154$
$(h + k) = 11$
$CA = r$
$(CA)^{2} = (r)^{2}$
$(h - 2)^{2} + (k - 2)^{2} = r^{2}$
$h^{2} + 4 - 4 h + k^{2} + 4 - 4 k = k^{2} (∵ r^{2} = k^{2})$
$h^{2} + 8 - 4 h - 4 (11 - h) = 0$
$h^{2} + 8 - 4 h - 44 + 4 h = 0$
$h^{2} = 36$
$h = \pm 6$
$P \equiv (\pm 6, 0), O \equiv (0, 0)$
$OP = 6 units$
original image

APEAPCET- 23.08.2021

Conic Section

119759 Find the equation of the circle which passes through origin and cuts off the intercepts - 2 and 3 over the $x$ and $y$ axes respectively.

1

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

2

2 (x^{2} + y^{2}) + 2 x - 3 y = 0

3

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

4

x^{2} + y^{2} + 2 x - 3 y = 0

Explanation:

D Given, $x intercept = - 2, y intercept = 3$
Let the circle be
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
$(0, 0), (- 2, 0), (0, 3)$
$(0, 0) \to c = 0$
$(- 2, 0) \to 4 - 4 g = 0$
$g = 1$
$(0, 3) \to 9 + 6 f = 0$
$6 f = - 9$
$f = \frac{- 9}{6} = - 3 / 2$
Required equation of circle
$x^{2} + y^{2} + 2 x - 3 y = 0$

AP EAPCET-25.08.2021

Conic Section

119760 The equation of the circle passing through $(0$ , $0)$ and which makes intercepts $a$ and $b$ on the coordinate axes is

1

x^{2} + y^{2} + a x + b y = 0

2

x^{2} + y^{2} - a x - b y = 0

3

x^{2} + y^{2} - a x + b y = 0

4

x^{2} + y^{2} - a x - b x = 0

Explanation:

C $: Centre = (\frac{a}{2}, \frac{b}{2})$
$Radius = \frac{\sqrt{a^{2} + b^{2}}}{2}$
$Equation of circle$
original image
${(x - \frac{a}{2})}^{2} + {(x - \frac{b}{2})}^{2} = \frac{a^{2} + b^{2}}{2}$
$4 x^{2} - 4 a x + a^{2} + 4 y^{2} - 4 b y + b^{2} = a^{2} + b^{2}$
$4 x^{2} + 4 y^{2} - 4 a x - 4 b y = 0$
$4 (x^{2} + y^{2} - a x - b y) = 0$
$x^{2} + y^{2} - a x - b y = 0$

AP EAMCET-17.09.2020

Conic Section

119756 If the lines $x + 2 y - 5 = 0$ and $3 x - y - 1 = 0$ denote two diameters of a circle of radius 5 units, then the equation of the circle is

1

x^{2} + y^{2} - 2 x + 4 y - 20 = 0

2

x^{2} + y^{2} - 2 x - 4 y - 20 = 0

3

x^{2} + y^{2} + 2 x - 4 y + 20 = 0

4

x^{2} + y^{2} + 2 x + 4 y + 20 = 0

Explanation:

B Equation of line are:-
$x + 2 y = 5$
$3 x - y = 1$
So,
$3 x + 6 y = 15$
$- 3 x + y = - 1$
Subtract (i) and (ii)
$\overset{―}{7 y} = 14 / 2$
$y = 2$
Put $y = 2$ in $3 x - y = 1$
$3 x - 2 = 1$
$3 x = 3$
$x = 1$
So, equation of circle is:-
$(x - 1)^{2} + (x - 2)^{2} = 5^{2}$
$x^{2} + 1 - 2 x + y^{2} + 4 - 4 y = 25$
$x^{2} + y^{2} - 2 x - 4 y - 20 = 0$

APEAPCET-20.08.2021

Conic Section

119757 Find the equation of a circle which cuts the circle $x^{2} + y^{2} - 6 x + 4 y - 3 = 0$ orthogonally while passing through $(3, 0)$ and touching the $y$ axis.

1

x^{2} + y^{2} + 6 x + 6 y + 9 = 0

2

x^{2} + y^{2} - 6 x - 6 y + 9 = 0

3

x^{2} + y^{2} - 6 x + 6 y - 9 = 0

4

x^{2} + y^{2} + 6 x - 6 y - 9 = 0

Explanation:

B Let, the equation of circle is:-
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
$Touches y-axis then f^{2} = c$
$Since (i) passes through (3, 0) then$
$9 + 6 g + c = 0$
$6 g + c = - 9$
$Given circle,$
$x^{2} + y^{2} - 6 x + 4 y - 3 = 0$
Then,
$g^{'} = - 3$
$f = 2$
$c^{'} = - 3$
Since (i) and (iii) are orthogonal then
$2 g^{'} + 2 f f = c + c^{'}$
$2 g (- 3) + 2 f (2) = c - 3$
$- 6 g + 4 f - c = - 3$
Then equation (iv) and (ii) add
$4 f = - 12$
$f = - 3$
Now,
$c = 9$
From equation (ii)
$6 g + 9 = - 9$
$6 g = - 18$
$g = 3$
$From (i) circle is$
$x^{2} + y^{2} - 6 x - 6 y + 9 = 0$ From (i) circle is

APEAPCET-23.08.2021

Conic Section

119758 Suppose a circle passes through $(2, 2)$ and $(9, 9)$ and touches the $x$ -axis at $P$ . If $O$ is the origin. then $O P$ is equal to

1 4

2 5

3 6

4 9

Explanation:

C
$r = \sqrt{(h - h)^{2} + (k - 0)^{2}}$
$r = \sqrt{k^{2}}$
$r^{2} = k^{2}$
$\Rightarrow CA = CB$
$\Rightarrow (CA)^{2} = (CB)^{2}$
$(h - 2)^{2} + (k - 2)^{2} = (h - 9)^{2} + (k - 9)^{2}$
$h^{2} + k^{2} + 4 + 4 - 4 h - 4 k = h^{2} + k^{2} + 81 + 81 - 18 h - 18 k$
$14 h + 14 k = 81 + 81 - 4 - 4$
$14 (h + k) = 154$
$(h + k) = 11$
$CA = r$
$(CA)^{2} = (r)^{2}$
$(h - 2)^{2} + (k - 2)^{2} = r^{2}$
$h^{2} + 4 - 4 h + k^{2} + 4 - 4 k = k^{2} (∵ r^{2} = k^{2})$
$h^{2} + 8 - 4 h - 4 (11 - h) = 0$
$h^{2} + 8 - 4 h - 44 + 4 h = 0$
$h^{2} = 36$
$h = \pm 6$
$P \equiv (\pm 6, 0), O \equiv (0, 0)$
$OP = 6 units$
original image

APEAPCET- 23.08.2021

Conic Section

119759 Find the equation of the circle which passes through origin and cuts off the intercepts - 2 and 3 over the $x$ and $y$ axes respectively.

1

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

2

2 (x^{2} + y^{2}) + 2 x - 3 y = 0

3

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

4

x^{2} + y^{2} + 2 x - 3 y = 0

Explanation:

D Given, $x intercept = - 2, y intercept = 3$
Let the circle be
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
$(0, 0), (- 2, 0), (0, 3)$
$(0, 0) \to c = 0$
$(- 2, 0) \to 4 - 4 g = 0$
$g = 1$
$(0, 3) \to 9 + 6 f = 0$
$6 f = - 9$
$f = \frac{- 9}{6} = - 3 / 2$
Required equation of circle
$x^{2} + y^{2} + 2 x - 3 y = 0$

AP EAPCET-25.08.2021

Conic Section

119760 The equation of the circle passing through $(0$ , $0)$ and which makes intercepts $a$ and $b$ on the coordinate axes is

1

x^{2} + y^{2} + a x + b y = 0

2

x^{2} + y^{2} - a x - b y = 0

3

x^{2} + y^{2} - a x + b y = 0

4

x^{2} + y^{2} - a x - b x = 0

Explanation:

C $: Centre = (\frac{a}{2}, \frac{b}{2})$
$Radius = \frac{\sqrt{a^{2} + b^{2}}}{2}$
$Equation of circle$
original image
${(x - \frac{a}{2})}^{2} + {(x - \frac{b}{2})}^{2} = \frac{a^{2} + b^{2}}{2}$
$4 x^{2} - 4 a x + a^{2} + 4 y^{2} - 4 b y + b^{2} = a^{2} + b^{2}$
$4 x^{2} + 4 y^{2} - 4 a x - 4 b y = 0$
$4 (x^{2} + y^{2} - a x - b y) = 0$
$x^{2} + y^{2} - a x - b y = 0$

AP EAMCET-17.09.2020

Conic Section

119756 If the lines $x + 2 y - 5 = 0$ and $3 x - y - 1 = 0$ denote two diameters of a circle of radius 5 units, then the equation of the circle is

1

x^{2} + y^{2} - 2 x + 4 y - 20 = 0

2

x^{2} + y^{2} - 2 x - 4 y - 20 = 0

3

x^{2} + y^{2} + 2 x - 4 y + 20 = 0

4

x^{2} + y^{2} + 2 x + 4 y + 20 = 0

Explanation:

B Equation of line are:-
$x + 2 y = 5$
$3 x - y = 1$
So,
$3 x + 6 y = 15$
$- 3 x + y = - 1$
Subtract (i) and (ii)
$\overset{―}{7 y} = 14 / 2$
$y = 2$
Put $y = 2$ in $3 x - y = 1$
$3 x - 2 = 1$
$3 x = 3$
$x = 1$
So, equation of circle is:-
$(x - 1)^{2} + (x - 2)^{2} = 5^{2}$
$x^{2} + 1 - 2 x + y^{2} + 4 - 4 y = 25$
$x^{2} + y^{2} - 2 x - 4 y - 20 = 0$

APEAPCET-20.08.2021

Conic Section

119757 Find the equation of a circle which cuts the circle $x^{2} + y^{2} - 6 x + 4 y - 3 = 0$ orthogonally while passing through $(3, 0)$ and touching the $y$ axis.

1

x^{2} + y^{2} + 6 x + 6 y + 9 = 0

2

x^{2} + y^{2} - 6 x - 6 y + 9 = 0

3

x^{2} + y^{2} - 6 x + 6 y - 9 = 0

4

x^{2} + y^{2} + 6 x - 6 y - 9 = 0

Explanation:

B Let, the equation of circle is:-
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
$Touches y-axis then f^{2} = c$
$Since (i) passes through (3, 0) then$
$9 + 6 g + c = 0$
$6 g + c = - 9$
$Given circle,$
$x^{2} + y^{2} - 6 x + 4 y - 3 = 0$
Then,
$g^{'} = - 3$
$f = 2$
$c^{'} = - 3$
Since (i) and (iii) are orthogonal then
$2 g^{'} + 2 f f = c + c^{'}$
$2 g (- 3) + 2 f (2) = c - 3$
$- 6 g + 4 f - c = - 3$
Then equation (iv) and (ii) add
$4 f = - 12$
$f = - 3$
Now,
$c = 9$
From equation (ii)
$6 g + 9 = - 9$
$6 g = - 18$
$g = 3$
$From (i) circle is$
$x^{2} + y^{2} - 6 x - 6 y + 9 = 0$ From (i) circle is

APEAPCET-23.08.2021

Conic Section

119758 Suppose a circle passes through $(2, 2)$ and $(9, 9)$ and touches the $x$ -axis at $P$ . If $O$ is the origin. then $O P$ is equal to

1 4

2 5

3 6

4 9

Explanation:

C
$r = \sqrt{(h - h)^{2} + (k - 0)^{2}}$
$r = \sqrt{k^{2}}$
$r^{2} = k^{2}$
$\Rightarrow CA = CB$
$\Rightarrow (CA)^{2} = (CB)^{2}$
$(h - 2)^{2} + (k - 2)^{2} = (h - 9)^{2} + (k - 9)^{2}$
$h^{2} + k^{2} + 4 + 4 - 4 h - 4 k = h^{2} + k^{2} + 81 + 81 - 18 h - 18 k$
$14 h + 14 k = 81 + 81 - 4 - 4$
$14 (h + k) = 154$
$(h + k) = 11$
$CA = r$
$(CA)^{2} = (r)^{2}$
$(h - 2)^{2} + (k - 2)^{2} = r^{2}$
$h^{2} + 4 - 4 h + k^{2} + 4 - 4 k = k^{2} (∵ r^{2} = k^{2})$
$h^{2} + 8 - 4 h - 4 (11 - h) = 0$
$h^{2} + 8 - 4 h - 44 + 4 h = 0$
$h^{2} = 36$
$h = \pm 6$
$P \equiv (\pm 6, 0), O \equiv (0, 0)$
$OP = 6 units$
original image

APEAPCET- 23.08.2021

Conic Section

119759 Find the equation of the circle which passes through origin and cuts off the intercepts - 2 and 3 over the $x$ and $y$ axes respectively.

1

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

2

2 (x^{2} + y^{2}) + 2 x - 3 y = 0

3

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

4

x^{2} + y^{2} + 2 x - 3 y = 0

Explanation:

D Given, $x intercept = - 2, y intercept = 3$
Let the circle be
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
$(0, 0), (- 2, 0), (0, 3)$
$(0, 0) \to c = 0$
$(- 2, 0) \to 4 - 4 g = 0$
$g = 1$
$(0, 3) \to 9 + 6 f = 0$
$6 f = - 9$
$f = \frac{- 9}{6} = - 3 / 2$
Required equation of circle
$x^{2} + y^{2} + 2 x - 3 y = 0$

AP EAPCET-25.08.2021

Conic Section

119760 The equation of the circle passing through $(0$ , $0)$ and which makes intercepts $a$ and $b$ on the coordinate axes is

1

x^{2} + y^{2} + a x + b y = 0

2

x^{2} + y^{2} - a x - b y = 0

3

x^{2} + y^{2} - a x + b y = 0

4

x^{2} + y^{2} - a x - b x = 0

Explanation:

C $: Centre = (\frac{a}{2}, \frac{b}{2})$
$Radius = \frac{\sqrt{a^{2} + b^{2}}}{2}$
$Equation of circle$
original image
${(x - \frac{a}{2})}^{2} + {(x - \frac{b}{2})}^{2} = \frac{a^{2} + b^{2}}{2}$
$4 x^{2} - 4 a x + a^{2} + 4 y^{2} - 4 b y + b^{2} = a^{2} + b^{2}$
$4 x^{2} + 4 y^{2} - 4 a x - 4 b y = 0$
$4 (x^{2} + y^{2} - a x - b y) = 0$
$x^{2} + y^{2} - a x - b y = 0$

AP EAMCET-17.09.2020

Conic Section

119756 If the lines $x + 2 y - 5 = 0$ and $3 x - y - 1 = 0$ denote two diameters of a circle of radius 5 units, then the equation of the circle is

1

x^{2} + y^{2} - 2 x + 4 y - 20 = 0

2

x^{2} + y^{2} - 2 x - 4 y - 20 = 0

3

x^{2} + y^{2} + 2 x - 4 y + 20 = 0

4

x^{2} + y^{2} + 2 x + 4 y + 20 = 0

Explanation:

B Equation of line are:-
$x + 2 y = 5$
$3 x - y = 1$
So,
$3 x + 6 y = 15$
$- 3 x + y = - 1$
Subtract (i) and (ii)
$\overset{―}{7 y} = 14 / 2$
$y = 2$
Put $y = 2$ in $3 x - y = 1$
$3 x - 2 = 1$
$3 x = 3$
$x = 1$
So, equation of circle is:-
$(x - 1)^{2} + (x - 2)^{2} = 5^{2}$
$x^{2} + 1 - 2 x + y^{2} + 4 - 4 y = 25$
$x^{2} + y^{2} - 2 x - 4 y - 20 = 0$

APEAPCET-20.08.2021

Conic Section

119757 Find the equation of a circle which cuts the circle $x^{2} + y^{2} - 6 x + 4 y - 3 = 0$ orthogonally while passing through $(3, 0)$ and touching the $y$ axis.

1

x^{2} + y^{2} + 6 x + 6 y + 9 = 0

2

x^{2} + y^{2} - 6 x - 6 y + 9 = 0

3

x^{2} + y^{2} - 6 x + 6 y - 9 = 0

4

x^{2} + y^{2} + 6 x - 6 y - 9 = 0

Explanation:

B Let, the equation of circle is:-
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
$Touches y-axis then f^{2} = c$
$Since (i) passes through (3, 0) then$
$9 + 6 g + c = 0$
$6 g + c = - 9$
$Given circle,$
$x^{2} + y^{2} - 6 x + 4 y - 3 = 0$
Then,
$g^{'} = - 3$
$f = 2$
$c^{'} = - 3$
Since (i) and (iii) are orthogonal then
$2 g^{'} + 2 f f = c + c^{'}$
$2 g (- 3) + 2 f (2) = c - 3$
$- 6 g + 4 f - c = - 3$
Then equation (iv) and (ii) add
$4 f = - 12$
$f = - 3$
Now,
$c = 9$
From equation (ii)
$6 g + 9 = - 9$
$6 g = - 18$
$g = 3$
$From (i) circle is$
$x^{2} + y^{2} - 6 x - 6 y + 9 = 0$ From (i) circle is

APEAPCET-23.08.2021

Conic Section

119758 Suppose a circle passes through $(2, 2)$ and $(9, 9)$ and touches the $x$ -axis at $P$ . If $O$ is the origin. then $O P$ is equal to

1 4

2 5

3 6

4 9

Explanation:

C
$r = \sqrt{(h - h)^{2} + (k - 0)^{2}}$
$r = \sqrt{k^{2}}$
$r^{2} = k^{2}$
$\Rightarrow CA = CB$
$\Rightarrow (CA)^{2} = (CB)^{2}$
$(h - 2)^{2} + (k - 2)^{2} = (h - 9)^{2} + (k - 9)^{2}$
$h^{2} + k^{2} + 4 + 4 - 4 h - 4 k = h^{2} + k^{2} + 81 + 81 - 18 h - 18 k$
$14 h + 14 k = 81 + 81 - 4 - 4$
$14 (h + k) = 154$
$(h + k) = 11$
$CA = r$
$(CA)^{2} = (r)^{2}$
$(h - 2)^{2} + (k - 2)^{2} = r^{2}$
$h^{2} + 4 - 4 h + k^{2} + 4 - 4 k = k^{2} (∵ r^{2} = k^{2})$
$h^{2} + 8 - 4 h - 4 (11 - h) = 0$
$h^{2} + 8 - 4 h - 44 + 4 h = 0$
$h^{2} = 36$
$h = \pm 6$
$P \equiv (\pm 6, 0), O \equiv (0, 0)$
$OP = 6 units$
original image

APEAPCET- 23.08.2021

Conic Section

119759 Find the equation of the circle which passes through origin and cuts off the intercepts - 2 and 3 over the $x$ and $y$ axes respectively.

1

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

2

2 (x^{2} + y^{2}) + 2 x - 3 y = 0

3

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

4

x^{2} + y^{2} + 2 x - 3 y = 0

Explanation:

D Given, $x intercept = - 2, y intercept = 3$
Let the circle be
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
$(0, 0), (- 2, 0), (0, 3)$
$(0, 0) \to c = 0$
$(- 2, 0) \to 4 - 4 g = 0$
$g = 1$
$(0, 3) \to 9 + 6 f = 0$
$6 f = - 9$
$f = \frac{- 9}{6} = - 3 / 2$
Required equation of circle
$x^{2} + y^{2} + 2 x - 3 y = 0$

AP EAPCET-25.08.2021

Conic Section

119760 The equation of the circle passing through $(0$ , $0)$ and which makes intercepts $a$ and $b$ on the coordinate axes is

1

x^{2} + y^{2} + a x + b y = 0

2

x^{2} + y^{2} - a x - b y = 0

3

x^{2} + y^{2} - a x + b y = 0

4

x^{2} + y^{2} - a x - b x = 0

Explanation:

C $: Centre = (\frac{a}{2}, \frac{b}{2})$
$Radius = \frac{\sqrt{a^{2} + b^{2}}}{2}$
$Equation of circle$
original image
${(x - \frac{a}{2})}^{2} + {(x - \frac{b}{2})}^{2} = \frac{a^{2} + b^{2}}{2}$
$4 x^{2} - 4 a x + a^{2} + 4 y^{2} - 4 b y + b^{2} = a^{2} + b^{2}$
$4 x^{2} + 4 y^{2} - 4 a x - 4 b y = 0$
$4 (x^{2} + y^{2} - a x - b y) = 0$
$x^{2} + y^{2} - a x - b y = 0$

AP EAMCET-17.09.2020