SATHEE: Circle 2 Question 5

Circle 2 Question 5

6. If one of the diameters of the circle, given by the equation, $x^{2} + y^{2} - 4 x + 6 y - 12 = 0$ , is a chord of a circle $S$ , whose centre is at $(- 3, 2)$ , then the radius of $S$ is

(a) $5 \sqrt{2}$

(b) $5 \sqrt{3}$

(d) 10

(2016 Main)

Show Answer

Answer:

Correct Answer: 6. (b)

Solution:

Given equation of a circle is $x^{2} + y^{2} - 4 x + 6 y - 12 = 0$ , whose centre is $(2, - 3)$ and radius

$= \sqrt{2^{2} + (- 3)^{2} + 12} = \sqrt{4 + 9 + 12} = 5$

Now, according to given information, we have the following figure.

$x^{2} + y^{2} - 4 x + 6 y - 12 = 0$

Clearly, $A O ⊥ B C$ , as $O$ is mid-point of the chord.

Now, in $△ A O B$ , we have

and

$\begin{aligned} O A & = \sqrt{(- 3 - 2)^{2} + (2 + 3)^{2}} \\ = \sqrt{25 + 25} = \sqrt{50} = 5 \sqrt{2} \end{aligned}$

$∴ A B = \sqrt{O A^{2} + O B^{2}} = \sqrt{50 + 25} = \sqrt{75} = 5 \sqrt{3}$

Play Store

App Store

Ask SATHEE

Welcome to SATHEE !
Select from 'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! 🌐📚🚀🎓

I'm relatively new and can sometimes make mistakes.
If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning.
To begin your journey now, click on "I understand".

Circle 2 Question 5

6. If one of the diameters of the circle, given by the equation, x2+y2−4x+6y−12=0, is a chord of a circle S, whose centre is at (−3,2), then the radius of S is

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu

6. If one of the diameters of the circle, given by the equation, $x^{2} + y^{2} - 4 x + 6 y - 12 = 0$ , is a chord of a circle $S$ , whose centre is at $(- 3, 2)$ , then the radius of $S$ is