If one of the diameters of the circle x^2+y^2-10 x+4 y+13=0 is a chord of another circle C , whose center is the point of intersection of the lines 2 x+3 y=12 and 3 x-2 y=5 , then the radius of the circle C is :

Circle PYQ — JEE Main 2024 · 31 Jan · Shift 1 · Q49 with Solution

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Correct answer: C

Find the center and radius of the given circle

The given circle is $x^2+y^2-10x+4y+13=0.$

Compare with the standard form: $x^2+y^2+2gx+2fy+c=0.$

So, $2g=-10 \Rightarrow g=-5, \qquad 2f=4 \Rightarrow f=2, \qquad c=13.$

Hence the center is $(-g,-f)=(5,-2).$

Its radius is

\sqrt{(-5)^2+2^2-13}= \sqrt{25+4-13}= \sqrt{16}=4.$$ So, the given circle has: - center $O=(5,-2)$ - radius $4$ Therefore, **any diameter** of this circle has length $$2r=8.$$ --- 2. **Find the center of the other circle $C$** The center of circle $C$ is the intersection of $$2x+3y=12$$ and $$3x-2y=5.$$ Solve these equations: From $$2x+3y=12 \quad ...(1)$$ $$3x-2y=5 \quad ...(2)$$ Multiply (1) by 2: $$4x+6y=24$$ Multiply (2) by 3: $$9x-6y=15$$ Add: $$13x=39 \Rightarrow x=3.$$ Substitute into (1): $$2(3)+3y=12 \Rightarrow 6+3y=12 \Rightarrow 3y=6 \Rightarrow y=2.$$ So the center of circle $C$ is $$P=(3,2).$$ --- 3. **Use the fact that a diameter of the first circle is a chord of circle $C$** A diameter of the first circle is a chord of circle $C$ of length $8$. For a circle, if a chord has length $L$ and its perpendicular distance from the center is $d$, then $$L=2\sqrt{R^2-d^2},$$ where $R$ is the radius of the circle. So we need the distance from the center $P=(3,2)$ of circle $C$ to that chord. Since the chord is a **diameter of the first circle**, it must pass through the center $O=(5,-2)$ of the first circle. To determine the radius uniquely from the options, note that the relevant diameter-chord must be the one whose midpoint is $O$, and the distance from $P$ to this chord is determined by the distance $OP$ when the perpendicular from $P$ to the chord passes through $O$. Compute: $$OP=\sqrt{(5-3)^2+(-2-2)^2}= \sqrt{2^2+(-4)^2}= \sqrt{4+16}=\sqrt{20}.$$ Thus, $$d=\sqrt{20}.$$ Now using chord length $8$: $$8=2\sqrt{R^2-d^2}$$ $$4=\sqrt{R^2-20}$$ $$16=R^2-20$$ $$R^2=36$$ $$R=6.$$ --- 4. **Check options** - A: $4$ - B: $3\sqrt2$ - C: $6$ ✅ - D: $\sqrt{20}$ So the radius of circle $C$ is $$\boxed{6}.$$ --- 5. **Comparison with stored answer** Stored correct answer: **C** Our derived answer: **C** They agree.

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View written solutionFree

Correct answer: C

Find the center and radius of the given circle

The given circle is $x^2+y^2-10x+4y+13=0.$

Compare with the standard form: $x^2+y^2+2gx+2fy+c=0.$

So, $2g=-10 \Rightarrow g=-5, \qquad 2f=4 \Rightarrow f=2, \qquad c=13.$

Hence the center is $(-g,-f)=(5,-2).$

Its radius is

\sqrt{(-5)^2+2^2-13}= \sqrt{25+4-13}= \sqrt{16}=4.$$ So, the given circle has: - center $O=(5,-2)$ - radius $4$ Therefore, **any diameter** of this circle has length $$2r=8.$$ --- 2. **Find the center of the other circle $C$** The center of circle $C$ is the intersection of $$2x+3y=12$$ and $$3x-2y=5.$$ Solve these equations: From $$2x+3y=12 \quad ...(1)$$ $$3x-2y=5 \quad ...(2)$$ Multiply (1) by 2: $$4x+6y=24$$ Multiply (2) by 3: $$9x-6y=15$$ Add: $$13x=39 \Rightarrow x=3.$$ Substitute into (1): $$2(3)+3y=12 \Rightarrow 6+3y=12 \Rightarrow 3y=6 \Rightarrow y=2.$$ So the center of circle $C$ is $$P=(3,2).$$ --- 3. **Use the fact that a diameter of the first circle is a chord of circle $C$** A diameter of the first circle is a chord of circle $C$ of length $8$. For a circle, if a chord has length $L$ and its perpendicular distance from the center is $d$, then $$L=2\sqrt{R^2-d^2},$$ where $R$ is the radius of the circle. So we need the distance from the center $P=(3,2)$ of circle $C$ to that chord. Since the chord is a **diameter of the first circle**, it must pass through the center $O=(5,-2)$ of the first circle. To determine the radius uniquely from the options, note that the relevant diameter-chord must be the one whose midpoint is $O$, and the distance from $P$ to this chord is determined by the distance $OP$ when the perpendicular from $P$ to the chord passes through $O$. Compute: $$OP=\sqrt{(5-3)^2+(-2-2)^2}= \sqrt{2^2+(-4)^2}= \sqrt{4+16}=\sqrt{20}.$$ Thus, $$d=\sqrt{20}.$$ Now using chord length $8$: $$8=2\sqrt{R^2-d^2}$$ $$4=\sqrt{R^2-20}$$ $$16=R^2-20$$ $$R^2=36$$ $$R=6.$$ --- 4. **Check options** - A: $4$ - B: $3\sqrt2$ - C: $6$ ✅ - D: $\sqrt{20}$ So the radius of circle $C$ is $$\boxed{6}.$$ --- 5. **Comparison with stored answer** Stored correct answer: **C** Our derived answer: **C** They agree.

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