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This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Givencircleis=\left(3,-1\right)\\ x^2+y^2-2x-4y-20=0\\ 2g=-2\Rightarrow g=-1\\ 2f=-4\Rightarrow f=-2\\ \therefore Centre{C}_1=\left(1,2\right)\\ andradiusr=\sqrt[]{g^2+f^2-c}\\ =\sqrt[]{1+4+20}=5\\ Letthecentreoftherequiredcirclebe\left(h,k\right).\\ Clearly,Pisthemidof{C}_1{C}_2\\ \therefore 5=\frac{1+h}{2}\Rightarrow h=9\\ and5=\frac{2+k}{2}\Rightarrow k=8\\ Radiusoftherequiredcircle=5\\ \therefore Equationofthecircleis\\ {\left(x-9\right)}^2+{\left(y-8\right)}^2={\left(5\right)}^2\\ \Rightarrow x^2+81-18x+y^2+64-16y=25\\ \Rightarrow x^2+y^2-18x-16y+145-25=0\\ \Rightarrow x^2+y^2-18x-16y+120=0\\ Hence,therequiredequationis{x}^2+y^2-18x-16y+120=0.\end{array}$

Seacom Skills University admissions 2025-26 are currently open for BTech and various courses. Students seeking admissiont to SSU Bolpur can visit the official website to apply online for their favorite courses. In the IIRF 2024 Ranking, Seacom University Bolpur was ranked 131 in the 'Overall' category. BTech admissions at SSU Bolpur are conducted through the WBJEE and JEE Mains entrance examinations followed by counselling.

Indian School of Business Management and Administration, Chennai admissions are ongoing. Candidates are required to participate in the admission process, which has been set by the institute. To apply for admissions candidates must follow the steps below: 

Step 1: Fill out the application form on the official website of the university. 

Step 2: Appear for the entrance exam if required. 

Step 3: Pay the course fee and secure your seat. 

Spicer Adventist University offers admission to candidates in the MBA programme based on their CET score. Aspirants with a valid score in PERA-CET or MAH MBA CET are offered admission in the MBA programme. Besides, candidates must also secure at least 50% aggregate in bachelor's programme for admission in MBA. Once selected, aspirants have to report at the campus for document verification and fee payment. 

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhat:\\ Centreofthecircle=\left(3,-1\right)\\ LengthofchordAB=6units\\ CP=\left|\frac{2\times 3-5\times 1+18}{\sqrt[]{{\left(2\right)}^2+{\left(-5\right)}^2}}\right|\\ =\left|\frac{6+5+18}{\sqrt[]{29}}\right|=\sqrt[]{29}\\ NowAB=6units.\\ \therefore AP=\frac{1}{2}AB=\frac{1}{2}\times 6=3units\\ In\Delta CPA,A{C}^2=C{P}^2+A{P}^2\\ ={\left(\sqrt[]{29}\right)}^2+{\left(3\right)}^2=29+9=38\\ \therefore AC=\sqrt[]{38}\\ So,theradiusofthecircle,r=\sqrt[]{38}\\ \therefore Equationofthecircleis\\ {\left(x-3\right)}^2+{\left(y+1\right)}^2={\left(\sqrt[]{38}\right)}^2\\ \Rightarrow {\left(x-3\right)}^2+{\left(y+1\right)}^2=38\\ \Rightarrow x^2+9-6x+y^2+1+2y=38\\ \Rightarrow x^2+y^2-6x+2y=28\\ Hence,therequiredequationis{x}^2+y^2-6x+2y=28.\end{array}$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Lettheequationofthecircleis\\ {\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\dots \left(i\right)\\ Ifthecircle\text{\hspace{0.17em}passes}through\left(2,3\right)and\left(4,5\right)then\\ {\left(2-h\right)}^2+{\left(3-k\right)}^2=r^2\dots \left(ii\right)\\ and{\left(4-h\right)}^2+{\left(5-k\right)}^2=r^2\dots \left(iii\right)\\ Subtractingeqn.\left(iii\right)fromeqn.\left(ii\right)wehave\\ {\left(2-h\right)}^2-{\left(4-h\right)}^2+{\left(3-k\right)}^2-{\left(5-k\right)}^2=0\\ \Rightarrow 4+h^2-4h-16-h^2+8h+9+k^2-6k-25-k^2+10k=0\\ \Rightarrow 4h+4k-28=0\\ \Rightarrow h+k=7\dots \left(iv\right)\\ ,thecentre\left(h,k\right)liesontheliney-4x+3=0\\ thenk-4h-3=0\Rightarrow k=4h-3\\ Puttingthevalueofkineqn.\left(iv\right)weget\\ h+4h-3=7\Rightarrow 5h=10\Rightarrow h=2\\ From\left(iv\right)wegetk=5\\ Puttingthevalueofhandkineqn.\left(ii\right)weget\\ {\left(2-2\right)}^2+{\left(3-5\right)}^2=r^2\Rightarrow r^2=4\\ So,theequationofthecircleis\\ {\left(x-2\right)}^2+{\left(y-5\right)}^2=4\\ \Rightarrow x^2+4-4x+y^2+25-10y=4\\ \Rightarrow x^2+y^2-4x-10y+25=0\\ Hence,therequiredequationis{x}^2+y^2-4x-10y+25=0.\end{array}$

The application forms for NID is usually released in September every year. The application forms are available online at the NID admission portal. Aspirants seeking admission in NID Gandhinagar have to register on the admission portal before the deadline. The applications usually closes by December. 

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Weknowthatthe\text{\hspace{0.17em}intersection point}ofthediametergivesthecentreofthecircle.\\ Givenequationsofdiametersare\\ 2x-3y=5\dots \left(i\right)\\ 3x-4y=7\dots \left(ii\right)\\ Fromeqn.\left(i\right)wehavex=\frac{5+3y}{2}\dots \left(iii\right)\\ Puttingthevalueofxineqn.\left(ii\right)wehave\\ 3\left(\frac{5+3y}{2}\right)-4y=7\\ \Rightarrow 15+9y-8y=14\Rightarrow y=14-15\Rightarrow y=-1\\ Nowfromeqn.\left(iii\right)wehavex=\frac{5+3\left(-1\right)}{2}\Rightarrow x=\frac{5-3}{2}\Rightarrow x=1\\ So,thecentreofthecircle=\left(1,-1\right)\\ Giventhatareaofthecircle=154\\ \Rightarrow \pi r^2=154\\ \Rightarrow \frac{22}{7}\times r^2=154\Rightarrow r^2=154\times \frac{7}{22}\\ \Rightarrow r^2=7\times 7\Rightarrow r=7\\ So,theequationofthecircleis\\ {\left(x-1\right)}^2+{\left(y+1\right)}^2={\left(7\right)}^2\\ \Rightarrow x^2+1-2x+y^2+1+2y=49\\ \Rightarrow x^2+y^2-2x+2y=47\\ Hence,therequiredequationofthecircleis{x}^2+y^2-2x+2y=47.\end{array}$

Bennett University admission is open for MBA and other courses. Students should visit the official website to know more and apply online. MBA course students can choose from various specialisations depending upon thir choice. 

Uma Krishna Shetty Institute of Management Studies and Research is a renowned private educational college and is located in Mumbai, Maharashtra.  The college has a state-of-the-art infrastructure with a 2000-seater auditorium, two 500-seater mini auditoriums, hostel facilities for boys and girls, banking facilities, high-tech air conditioning, and Wi-Fi-enabled classrooms.