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Byju's Answer

Standard XII

Mathematics

Equation of Tangent in Slope Form

Let C1 and ...

Question

Let $C_{1}$ and $C_{2}$ are circles defined by $x^{2} + y^{2} - 20 x + 64 = 0$ and $x^{2} + y^{2} + 30 x + 144 = 0$ . The length of the shortest line segment PQ that is tangent to $C_{1}$ at P and $C_{2}$ at Q is

15

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20

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Solution

The correct option is C $20$

$C_{1} = x^{2} + y^{2} - 20 x + 64 = 0$
$C_{1} = x^{2} + y^{2} + 30 x + 144 = 0$
$C_{1} : (10, 0), r_{1} = 6$ \\ $C_{2} : (- 15, 0), r_{2} = 9$
$C_{1} C_{2} = 25, r_{1} + r_{2} = 15$
$C_{1} C_{2} > r_{1} + r_{2}$
$m : n = 9 : 6$
$R = (\frac{9 \times 10}{15} - \frac{6 (- 15)}{2}) = (60, 0)$
$Q S + P S = \sqrt{15^{2} {- 9}^{2}} + \sqrt{10^{2} {- 6}^{2}} = 12 + 8 = 20$

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Similar questions

Q. If

C_{1}

and

C_{2}

are circles whose equations are

x^{2} + y^{2} - 20 x + 64 = 0

and

x^{2} + y^{2} + 30 x + 144 = 0

, then the length of the shortest line segment

P Q

that is tangent to

C_{1}

P

and to

C_{2}

Q

If $c_{1} : x^{2} + y^{2} - 20 x + 64 = 0$ and $c_{2} :$ $x^{2} + y^{2} + 30 x + 144 = 0$ . Then the length of the shortest line segment $P Q$ which is tangent to $c_{1}$ at $P$ and $c_{2}$ at $Q$ is

Q. If

C_{1} : x^{2} + y^{2} - 20 x + 64 = 0

and

C_{2} : x^{2} + y^{2} + 30 x + 144 = 0

. The line PQ touches the circle touches

C_{1}

P

and

C_{2}

Q

. Then the length of the shortest line segment

P Q

Q. If

C_{1} : x^{2} + y^{2} - 20 x + 64 = 0

and

C_{2} : x^{2} + y^{2} + 30 x + 144 = 0

. Then the length of the shortest line segment

P Q

which touches

C_{1}

P

and to

C_{2}

Q

is ?

Q. Let

L_{1}, L_{2}

and

L_{3}

be the lengths of tangents drawn from a point

P (h, k)

to the circles

x^{2} + y^{2} = 4,

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

and

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

respectively. Let

L_{1}^{4} = L_{2}^{2} L_{3}^{2} + 16

gives the locus of

P

as two curves,

C_{1}

(straight line) and

C_{2}

(circle). A triangle is formed by

C_{1}

and two other lines which are at an angle of

45^{\circ}

with

C_{1}

and tangent to the circle

C_{2} .

Then which of the following is/are correct?

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Let C1 and C2 are circles defined by x2+y2−20x+64=0 and x2+y2+30x+144=0. The length of the shortest line segment PQ that is tangent to C1 at P and C2 at Q is

The correct option is C 20C1=x2+y2−20x+64=0C1=x2+y2+30x+144=0C1:(10,0),r1=6\\ C2:(−15,0),r2=9C1C2=25,r1+r2=15C1C2>r1+r2m:n=9:6R=(9×1015−6(−15)2)=(60,0)QS+PS=√152−92+√102−62=12+8=20

Let $C_{1}$ and $C_{2}$ are circles defined by $x^{2} + y^{2} - 20 x + 64 = 0$ and $x^{2} + y^{2} + 30 x + 144 = 0$ . The length of the shortest line segment PQ that is tangent to $C_{1}$ at P and $C_{2}$ at Q is